Rd Sharma XII Vol 2 2019 for Class 12 Commerce Math Chapter 1 - Definite Integrals

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Page No 20.110:

Question 1:

$\int_{0}^{3} (x + 4) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . + f (a + (n - 1) h)], where, h = \frac{b - a}{n}$

$Here, a = 0, b = 3, f (x) = x + 4, h = \frac{3 - 0}{n} = \frac{3}{n} Therefore, I = \int_{0}^{3} (x + 4) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [(0 + 4) + (h + 4) + . . . . . . . + ((n - 1) h + 4)] = \lim_{h \to 0} h [4 n + h (1 + 2 + . . . . . . . + (n - 1))] = \lim_{h \to 0} h [4 n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [4 n + \frac{3}{n} \times \frac{n (n - 1)}{2}] = \lim_{n \to \infty} [12 + \frac{9}{2} (1 - \frac{1}{n})] = 12 + \frac{9}{2} = \frac{33}{2}$

Page No 20.110:

Question 2:

$\int_{0}^{2} (x + 3) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x + 3, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x + 3) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [(0 + 3) + (0 + h + 3) + . . . . . . . . . . . . . . . + (0 + (n - 1) h + 3)] = \lim_{h \to 0} h [3 n + h \{1 + 2 + 3 . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [3 n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [3 n + n - 1] = \lim_{n \to \infty} 2 (4 - \frac{1}{n}) = 8$

Page No 20.110:

Question 3:

$\int_{1}^{3} (3 x - 2) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = 1, b = 3, f (x) = 3 x - 2, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, I = \int_{1}^{3} (3 x - 2) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(3 - 2) + (3 + 3 h - 2) + (3 + 6 h - 2) . . . . . . . . . . . . . . . + (3 (n - 1) h + 3 - 2)] = \lim_{h \to 0} h [n + 3 h (1 + 2 + 3 . . . . . . . . . + (n - 1))] = \lim_{h \to 0} h [n + 3 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [n + 3 n - 3] = \lim_{n \to \infty} 2 (4 - \frac{3}{n}) = 8$

Page No 20.110:

Question 4:

$\int_{- 1}^{1} (x + 3) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = - 1, b = 1, f (x) = x + 3, h = \frac{1 + 1}{n} = \frac{2}{n} Therefore, I = \int_{- 1}^{1} (x + 3) d x = \lim_{h \to 0} h [f (- 1) + f (- 1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{- 1 + (n - 1) h\}] = \lim_{h \to 0} h [(- 1 + 3) + (- 1 + h + 3) + . . . . . . . . . . . . . . . + \{- 1 + (n - 1) h + 3\}] = \lim_{h \to 0} h [2 n + h \{1 + 2 + 3 . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [2 n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [2 n + n - 1] = \lim_{n \to \infty} 2 (3 - \frac{1}{n}) = 6$

Page No 20.110:

Question 5:

$\int_{0}^{5} (x + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here, a = 0, b = 5, f (x) = x + 1, h = \frac{5 - 0}{n} = \frac{5}{n} Therefore, I = \int_{0}^{5} (x + 1) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 1) + (h + 1) + . . . . . . . . . . . . . . . + \{(n - 1) h + 1\}] = \lim_{h \to 0} h [n + h \{1 + 2 + 3 . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{5}{n} [n + \frac{5 n - 5}{2}] = \lim_{n \to \infty} 5 (\frac{7}{2} - \frac{5}{n}) = \frac{35}{2}$

Page No 20.110:

Question 6:

$\int_{1}^{3} (2 x + 3) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here, a = 1, b = 3, f (x) = 2 x + 3, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, I = \int_{1}^{3} (2 x + 3) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(2 + 3) + (2 + 2 h + 3) + . . . . . . . . . . . . . . . + \{2 + 2 (n - 1) h + 3\}] = \lim_{h \to 0} h [5 n + 2 h \{1 + 2 + 3 . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [5 n + 2 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [5 n + 2 n - 2] = \lim_{n \to \infty} 2 (7 - \frac{2}{n}) = 14$

Page No 20.110:

Question 7:

$\int_{3}^{5} (2 - x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 3, b = 5, f (x) = 2 - x, h = \frac{5 - 3}{n} = \frac{2}{n} Therefore, I = \int_{3}^{5} (2 - x) d x = \lim_{h \to 0} h [f (2) + f (2 + h) + . . . + f (2 + (n - 1) h)] = \lim_{h \to 0} h [(2 - 2) + (2 - h - 2) + . . . + (2 - (n - 1) h - 2)] = \lim_{h \to 0} h [- h (1 + 2 + 3 + . . . + (n - 1))] = \lim_{h \to 0} h [- 2 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [- 2 n + 2] = \lim_{n \to \infty} 2 (- 2 + \frac{2}{n}) = - 4$

Page No 20.110:

Question 8:

$\int_{0}^{2} (x^{2} + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + 1, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + 1) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [(0 + 1) + (h^{2} + 1) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + 1\}] = \lim_{h \to 0} h [n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{2}{n} [n + \frac{2 (n - 1) (2 n - 1)}{3 n}] = \lim_{n \to \infty} 2 \{1 + \frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = 2 + \frac{8}{3} = \frac{14}{3}$

Page No 20.110:

Question 9:

$\int_{1}^{2} x^{2} d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 1, b = 2, f (x) = x^{2}, h = \frac{2 - 1}{n} = \frac{1}{n} Therefore, I = \int_{1}^{2} (x^{2}) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(1) + {(h + 1)}^{2} + . . . . . . . . . . . . . . . + {((n - 1) h + 1)}^{2}] = \lim_{h \to 0} h [n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 2 h \{1 + 2 + 3 + . . . . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 2 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{1}{n} [n + \frac{(n - 1) (2 n - 1)}{6 n} + n - 1] = \lim_{n \to \infty} \{2 + \frac{1}{6} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{1}{n}\} = 2 + \frac{1}{3} = \frac{7}{3}$

Page No 20.110:

Question 10:

$\int_{2}^{3} (2 x^{2} + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 2, b = 3, f (x) = 2 x^{2} + 1, h = \frac{3 - 2}{n} = \frac{1}{n} Therefore, I = \int_{2}^{3} (2 x^{2} + 1) d x = \lim_{h \to 0} h [f (2) + f (2 + h) + . . . . . . . . . . . . . . . . . . . . + f \{2 + (n - 1) h\}] = \lim_{h \to 0} h [2 (2 . 2^{2}) + 1 + \{2 {(2 + h)}^{2} + 1\} + . . . . . . . . . . . . . . . + \{2 {((2 + n - 1) h)}^{2} + 1\}] = \lim_{h \to 0} h [n + 2 \{2^{2} + {(2 + h)}^{2} + . . . . . . . . . . . . . {((2 + n - 1) h)}^{2}\}] = \lim_{h \to 0} h [n + 8 n + 2 h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 8 h \{1 + 2 + . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [9 n + h^{2} \frac{2 n (n - 1) (2 n - 1)}{6} + 8 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{1}{n} [9 n + \frac{(n - 1) (2 n - 1)}{3 n} + 4 n - 4] = \lim_{n \to \infty} \{13 + \frac{1}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{4}{n}\} = 13 + \frac{2}{3} = \frac{41}{3}$

Page No 20.110:

Question 11:

$\int_{1}^{2} (x^{2} - 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 1, b = 2, f (x) = x^{2} - 1, h = \frac{2 - 1}{n} = \frac{1}{n} Therefore, I = \int_{1}^{2} (x^{2} - 1) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(1 - 1) + (h^{2} - 1) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} - 1\}] = \lim_{h \to 0} h [n - 1 + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [n - 1 + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{1}{n} [n - 1 + \frac{(n - 1) (2 n - 1)}{6 n}] = \lim_{n \to \infty} \{1 - \frac{1}{n} + \frac{1}{6} (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = 1 + \frac{1}{3} = \frac{4}{3}$

Page No 20.110:

Question 12:

$\int_{0}^{2} (x^{2} + 4) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + 4, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + 4) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 4) + (h^{2} + 4) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + 4\}] = \lim_{h \to 0} h [4 n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [4 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{2}{n} [4 n + \frac{2 (n - 1) (2 n - 1)}{3 n}] = \lim_{n \to \infty} 2 \{4 + \frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = 8 + \frac{8}{3} = \frac{32}{3}$

Page No 20.111:

Question 13:

$\int_{1}^{4} (x^{2} - x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 1, b = 4, f (x) = x^{2} - x, h = \frac{4 - 1}{n} = \frac{3}{n} Therefore, I = \int_{1}^{4} (x^{2} - x) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(1 - 1) + \{{(1 + h)}^{2} - (1 + h)\} + . . . . . . . . . . . . . . . + \{{(1 + (n - 1) h)}^{2} - (1 + (n - 1) h)\}] = \lim_{h \to 0} h [h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 1 + 2 h \{1 + 2 + . . . . . . + (n - 1)\} - n - h \{1 + 2 + . . . . . + (n - 1)\}] = \lim_{h \to 0} h [h^{2} \frac{n (n - 1) (2 n - 1)}{6} + h \frac{(n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [\frac{9 (n - 1) (2 n - 1)}{6 n} + \frac{3 (n - 1)}{2}] = \lim_{n \to \infty} 3 [\frac{3}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + \frac{3}{2} (1 - \frac{1}{n})] = 9 + \frac{9}{2} = \frac{27}{2}$

Page No 20.111:

Question 14:

$\int_{0}^{1} (3 x^{2} + 5 x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 1, f (x) = 3 x^{2} + 5 x, h = \frac{1 - 0}{n} = \frac{1}{n} Therefore, I = \int_{0}^{1} (3 x^{2} + 5 x) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 0) + (3 h^{2} + 5 h) + . . . . . . . . . . . . . . . + \{3 {(n - 1)}^{2} h^{2} + 5 (n - 1) h\}] = \lim_{h \to 0} h [5 h (1 + 2 + . . . . . . . . . + n) + 3 h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [5 h \frac{n (n - 1)}{2} + h^{2} \frac{3 n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{1}{n} [\frac{5 (n - 1)}{2} + \frac{(n - 1) (2 n - 1)}{2 n}] = \lim_{n \to \infty} [\frac{5}{2} (1 - \frac{1}{n}) + \frac{1}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n})] = \frac{5}{2} + 1 = \frac{7}{2}$

Page No 20.111:

Question 15:

$\int_{0}^{2} e^{x} d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = e^{x}, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} e^{x} d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [e^{0} + e^{h} + e^{2 h} + . . . . . . . + e^{(n - 1) h}] = \lim_{h \to 0} h [\frac{{(e^{h})}^{n} - 1}{e^{h} - 1}] = \lim_{h \to 0} [\frac{e^{2} - 1}{\frac{e^{h} - 1}{h}}] = \frac{e^{2} - 1}{1} = e^{2} - 1$

Page No 20.111:

Question 16:

$\int_{a}^{b} e^{x} d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = a, b = b, f (x) = e^{x}, h = \frac{b - a}{n} Therefore, I = \int_{a}^{b} e^{x} d x = \lim_{h \to 0} h [f (a) + f (a + h) + . . . . . . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] = \lim_{h \to 0} h [e^{a} + e^{a + h} + . . . . . . . . . . . . + e^{\{a + (n - 1) h\}}] = \lim_{h \to 0} h [e^{a} \{\frac{{(e^{h})}^{n} - 1}{e^{h} - 1}\}] = \lim_{h \to 0} h [e^{a} \frac{e^{b - a} - 1}{e^{h} - 1}] = \lim_{h \to 0} [\frac{e^{b} - e^{a}}{\frac{e^{h} - 1}{h}}] = \frac{e^{b} - e^{a}}{1} = e^{b} - e^{a}$

Page No 20.111:

Question 17:

$\int_{a}^{b} \cos x d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = a, b = b, f (x) = \cos x, h = \frac{b - a}{n} Therefore, I = \int_{a}^{b} \cos x d x = \lim_{h \to 0} h [f (a) + f (a + h) + . . . + f \{a + (n - 1) h\}] = \lim_{h \to 0} h [\cos (a) + \cos (a + h) + . . . + \cos \{a + (n - 1) h\}] = \lim_{h \to 0} h [\frac{\cos \{a + (n - 1) \frac{h}{2}\} \sin \frac{n h}{2}}{\sin \frac{h}{2}}] = \lim_{h \to 0} [\frac{\frac{h}{2}}{\sin \frac{h}{2}} 2 \cos (a + \frac{b - a}{2} - \frac{h}{2}) \sin (\frac{b - a}{2})] (Using n h = b - a) = \lim_{h \to 0} \frac{\frac{h}{2}}{\sin \frac{h}{2}} \times \lim_{h \to 0} 2 \cos (\frac{a + b}{2} - \frac{h}{2}) \sin (\frac{b - a}{2}) = 2 \cos (\frac{a + b}{2}) \sin (\frac{b - a}{2}) = \sin b - \sin a [Since, 2 \cos A \sin B = \sin (A + B) - \sin (A - B)]$

Page No 20.111:

Question 18:

$\int_{0}^{π / 2} \sin x d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = 0, b = \frac{π}{2}, f (x) = \sin x, h = \frac{\frac{π}{2} - 0}{n} = \frac{π}{2 n} Therefore, I = \int_{0}^{\frac{π}{2}} \sin x d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [\sin 0 + \sin h + \sin 2 h + . . . + \sin (n - 1) h] = \lim_{h \to 0} h [\frac{\sin ((n - 1) \frac{h}{2}) \sin \frac{n h}{2}}{\sin \frac{h}{2}}] = \lim_{h \to 0} [\frac{\frac{h}{2}}{\sin \frac{h}{2}} \times 2 \sin (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4}] (Using n h = \frac{π}{2}) = \lim_{h \to 0} \frac{\frac{h}{2}}{\sin \frac{h}{2}} \times \lim_{h \to 0} 2 \sin (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4} = 2 \sin \frac{π}{4} \sin \frac{π}{4} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1$

Page No 20.111:

Question 19:

$\int_{0}^{π / 2} \cos x d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$
$Here a = 0, b = \frac{π}{2}, f (x) = \cos x, h = \frac{\frac{π}{2} - 0}{n} = \frac{π}{2 n} Therefore, I = \int_{0}^{\frac{π}{2}} \cos x d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [\cos 0 + \cos h + \cos 2 h + . . . + \cos (n - 1) h] = \lim_{h \to 0} h [\frac{\cos ((n - 1) \frac{h}{2}) \sin \frac{n h}{2}}{\sin \frac{h}{2}}] = \lim_{h \to 0} h [\frac{\cos (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4}}{\sin \frac{h}{2}}] (Using, n h = \frac{π}{2}) = \lim_{h \to 0} [\frac{\frac{h}{2}}{\sin \frac{h}{2}} \times 2 \cos (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4}] = \lim_{h \to 0} \frac{\frac{h}{2}}{\sin \frac{h}{2}} \times \lim_{h \to 0} 2 \cos (\frac{π}{4} - \frac{h}{2}) \sin \frac{π}{4} = 2 \cos \frac{π}{4} \sin \frac{π}{4} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1$

Page No 20.111:

Question 20:

$\int_{1}^{4} (3 x^{2} + 2 x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f (a + (n - 1) h)] where h = \frac{b - a}{n}$

$Here a = 1, b = 4, f (x) = 3 x^{2} + 2 x, h = \frac{4 - 1}{n} = \frac{3}{n} Therefore, I = \int_{1}^{4} (3 x^{2} + 2 x) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(3 . 1^{2} + 2 \times 1) + (3 {(1 + h)}^{2} + 2 (1 + h)) + . . . . . . . . . . . . . . . + \{3 {(1 + (n - 1) h)}^{2} + 2 (1 + (n - 1) h)\}] = \lim_{h \to 0} h [3 \{1^{2} + {(1 + h)}^{2} + {(1 + 2 h)}^{2} + . . . . . . . . . . . + {(1 + (n - 1) h)}^{2}\} + 2 \{1 + (1 + h) + . . . . . . . . . . + (1 + (n - 1) h)\}] = \lim_{h \to 0} h [3 n + 3 h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 6 h \{1 + 2 + . . . . . . . . . (n - 1) h\} + 2 n + 2 h \{1 + 2 + . . . . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [5 n + 3 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 8 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [5 n + \frac{9 (n - 1) (2 n - 1)}{2 n} + 12 n - 12] = \lim_{n \to \infty} 3 [17 - \frac{12}{n} + \frac{9}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n})] = 51 + 27 = 78$

Page No 20.111:

Question 21:

$\int_{0}^{2} (3 x^{2} - 2) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = 3 x^{2} - 2, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (3 x^{2} - 2) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 - 2) + (3 h^{2} - 2) + . . . . . . . . . . . . . . . + \{3 {(n - 1)}^{2} h^{2} - 2\}] = \lim_{h \to 0} h [- 2 n + 3 h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [- 2 n + 3 h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{2}{n} [- 2 n + \frac{2 (n - 1) (2 n - 1)}{n}] = \lim_{n \to \infty} 2 \{- 2 + 2 (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = - 4 + 8 = 4$

Page No 20.111:

Question 22:

$\int_{0}^{2} (x^{2} + 2) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + 2, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + 2) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 2) + (h^{2} + 2) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + 2\}] = \lim_{h \to 0} h [2 n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\}] = \lim_{h \to 0} h [2 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to \infty} \frac{2}{n} [2 n + \frac{2 (n - 1) (2 n - 1)}{3 n}] = \lim_{n \to \infty} 2 \{2 + \frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})\} = 4 + \frac{8}{3} = \frac{20}{3}$

Page No 20.111:

Question 23:

$\int_{0}^{4} (x + e^{2 x}) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here, a = 0, b = 4, f (x) = x + e^{2 x}, h = \frac{4 - 0}{n} = \frac{4}{n} Therefore, I = \int_{0}^{4} (x + e^{2 x}) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + e^{0}) + (h + e^{2 h}) + . . . . . . . . . . . . . . . + \{(n - 1) h + e^{2 (n - 1) h}\}] = \lim_{h \to 0} h [h \{1 + 2 + . . . . . + (n - 1) h\} + e^{0} + e^{2 h} + e^{4 h} + . . . . . . . . . + e^{2 (n - 1) h}] = \lim_{h \to 0} h [h \frac{n (n - 1)}{2} + \frac{{(e^{2 h})}^{n} - 1}{e^{2 h} - 1}] = \lim_{n \to \infty} \frac{16}{n^{2}} \times \frac{n (n - 1)}{2} + \lim_{h \to 0} \frac{e^{8} - 1}{\frac{e^{2 h} - 1}{h}} = \lim_{n \to \infty} 8 (1 - \frac{1}{n}) + \lim_{h \to 0} \frac{e^{8} - 1}{\frac{2 (e^{2 h} - 1)}{2 h}} = 8 + \frac{e^{8} - 1}{2} = \frac{15 + e^{8}}{2}$

Page No 20.111:

Question 24:

$\int_{0}^{2} (x^{2} + x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + x, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + x) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 0) + (h^{2} + h) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + h\}] = \lim_{h \to 0} h [h^{2} (1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}) + h \{1 + 2 + 3 . . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [h^{2} \frac{n (n - 1) (2 n - 1)}{6} + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [\frac{2 (n - 1) (2 n - 1)}{3 n} + n - 1] = \lim_{n \to \infty} 2 [\frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + 1 - \frac{1}{n}] = \frac{8}{3} + 2 = \frac{14}{3}$

Page No 20.111:

Question 25:

$\int_{0}^{2} (x^{2} + 2 x + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} + 2 x + 1, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} + 2 x + 1) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 0 + 1) + (h^{2} + 2 h + 1) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} + 2 (n - 1) h + 1\}] = \lim_{h \to 0} h [n + h^{2} (1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}) + 2 h \{1 + 2 + . . . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 2 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [n + \frac{2 (n - 1) (2 n - 1)}{3 n} + 2 n - 2] = \lim_{n \to \infty} 2 \{3 + \frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{2}{n}\} = 6 + \frac{8}{3} = \frac{26}{3}$

Page No 20.111:

Question 26:

$\int_{0}^{3} (2 x^{2} + 3 x + 5) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 3, f (x) = 2 x^{2} + 3 x + 5, h = \frac{3 - 0}{n} = \frac{3}{n} Therefore, I = \int_{0}^{3} (2 x^{2} + 3 x + 5) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 0 + 5) + (2 h^{2} + 3 h + 5) + . . . . . . . . . . . . . . . + \{2 {(n - 1)}^{2} h^{2} + 3 (n - 1) h + 5\}] = \lim_{h \to 0} h [5 n + 2 h^{2} (1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}) + 3 h \{1 + 2 + . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [5 n + 2 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 3 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [5 n + \frac{3 (n - 1) (2 n - 1)}{n} + \frac{9 (n - 1)}{2}] = \lim_{n \to \infty} 3 [5 + 3 (1 - \frac{1}{n}) (2 - \frac{1}{n}) + \frac{9}{2} (1 - \frac{1}{n})] = 15 + 18 + \frac{27}{2} = \frac{93}{3}$

Page No 20.111:

Question 27:

$\int_{a}^{b} x d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = a, b = b, f (x) = x, h = \frac{b - a}{n} Therefore, I = \int_{a}^{b} x d x = \lim_{h \to 0} h [f (a) + f (a + h) + . . . . . . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] = \lim_{h \to 0} h [a + (a + h) + (a + 2 h) + . . . . . . . . . . + \{a + (n - 1) h\}] = \lim_{h \to 0} h [n a + h \{1 + 2 + 3 + . . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [n a + h \frac{n (n - 1)}{2}] = \lim_{h \to 0} \frac{b - a}{n} [n a + \frac{[b - a] (n - 1)}{2}] = \lim_{h \to 0} [(b - a) a + \frac{(b - a) (b - a - h)}{2}] = (b - a) a + \frac{{(b - a)}^{2}}{2} = \frac{2 a b - 2 a^{2} + b^{2} + a^{2} - 2 a b}{2} = \frac{b^{2} - a^{2}}{2}$

Page No 20.111:

Question 28:

$\int_{0}^{5} (x + 1) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 5, f (x) = x + 1, h = \frac{5 - 0}{n} = \frac{5}{n} Therefore, I = \int_{0}^{5} (x + 1) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 + 1) + (h + 1) + . . . . . . . . . . . . . . . + \{(n - 1) h + 1\}] = \lim_{h \to 0} h [n + h \{1 + 2 + 3 + . . . . . . . . . . . . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [n + h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{5}{n} [n + \frac{5 (n - 1)}{2}] = \lim_{n \to \infty} 5 [1 + \frac{5}{2} (1 - \frac{1}{n})] = 5 + \frac{25}{2} = \frac{35}{2}$

Page No 20.111:

Question 29:

$\int_{2}^{3} x^{2} d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 2, b = 3, f (x) = x^{2}, h = \frac{3 - 2}{n} = \frac{1}{n} Therefore, I = \int_{2}^{3} x^{2} d x = \lim_{h \to 0} h [f (2) + f (2 + h) + . . . . . . . . . . . . . . . . . . . . + f \{2 + (n - 1) h\}] = \lim_{h \to 0} h [2^{2} + {(2 + h)}^{2} + . . . . . . . . . . . + {\{2 + (n - 1) h\}}^{2}] = \lim_{h \to 0} h [4 n + h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} + 4 h \{1 + 2 + . . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [4 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 4 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{1}{n} [4 n + \frac{(n - 1) (2 n - 1)}{6 n} + 2 n - 2] = \lim_{n \to \infty} [6 + \frac{1}{6} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{2}{n}] = 6 + \frac{1}{3} = \frac{19}{3}$

Page No 20.111:

Question 30:

$\int_{1}^{4} (x^{2} - x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 1, b = 4, f (x) = x^{2} - x, h = \frac{4 - 1}{n} = \frac{3}{n} Therefore, I = \int_{1}^{4} (x^{2} - x) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(1 - 1) + {(1 + h)}^{2} - (1 + h) + . . . . . . . . . . . . . . . + {\{(n - 1) h + 1\}}^{2} - \{(n - 1) h + 1\}] = \lim_{h \to 0} h [h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} - h \{1 + 2 + . . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [h^{2} \frac{n (n - 1) (2 n - 1)}{6} - h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{3}{n} [\frac{3 (n - 1) (2 n - 1)}{2 n} + \frac{3 (n - 1)}{2}] = \lim_{n \to \infty} 3 [\frac{3}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + \frac{3}{2} (1 - \frac{1}{n})] = 9 + \frac{9}{3} = \frac{38}{3}$

Page No 20.111:

Question 31:

$\int_{0}^{2} (x^{2} - x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here a = 0, b = 2, f (x) = x^{2} - x, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, I = \int_{0}^{2} (x^{2} - x) d x = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . . . . . . . . . . . + f \{0 + (n - 1) h\}] = \lim_{h \to 0} h [(0 - 0) + (h^{2} - h) + . . . . . . . . . . . . . . . + \{{(n - 1)}^{2} h^{2} - (n - 1) h\}] = \lim_{h \to 0} h [h^{2} \{1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}\} - h \{1 + 2 + . . . . . + (n - 1) h\}] = \lim_{h \to 0} h [h^{2} \frac{n (n - 1) (2 n - 1)}{6} - h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [\frac{2 (n - 1) (2 n - 1)}{3 n} - n + 1] = \lim_{n \to \infty} 2 [\frac{2}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - 1 + \frac{1}{n}] = \frac{8}{3} - 2 = \frac{2}{3}$

Page No 20.111:

Question 32:

$\int_{1}^{3} (2 x^{2} + 5 x) d x$

Answer:

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) . . . . . . . . . . . . . . . + f \{a + (n - 1) h\}] where h = \frac{b - a}{n}$

$Here, a = 1, b = 3, f (x) = 2 x^{2} + 5 x, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, I = \int_{1}^{3} (2 x^{2} + 5 x) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . . . . . . . . . . . + f \{1 + (n - 1) h\}] = \lim_{h \to 0} h [(2 + 5) + \{2 {(1 + h)}^{2} + 5 (1 + h)\} + . . . . . . . . . . . . . . . + \{2 {(1 + (n - 1) h)}^{2} + 5 (1 + (n - 1) h)\}] = \lim_{h \to 0} h [2 \{1^{2} + {(1 + h)}^{2} + . . . . . . . . . . . . + {\{1 + (n - 1) h\}}^{2}\} + 5 \{1 + (1 + h) + (1 + 2 h + . . . . . . . . + (1 + (n - 1) h))\}] = \lim_{h \to 0} h [2 n + 2 h^{2} (1^{2} + 2^{2} + 3^{2} . . . . . . . . . + {(n - 1)}^{2}) + 4 h \{1 + 2 + . . . . . . + (n - 1)\} + 5 n + 5 h \{1 + 2 + . . . . . . + (n - 1)\}] = \lim_{h \to 0} h [7 n + 2 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 9 h \frac{n (n - 1)}{2}] = \lim_{n \to \infty} \frac{2}{n} [7 n + \frac{4 (n - 1) (2 n - 1)}{3 n} + 9 n - 9] = \lim_{n \to \infty} 2 [16 + \frac{4}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) - \frac{9}{n}] = 32 + \frac{16}{3} = \frac{112}{3}$

Page No 20.111:

Question 33:

Evaluate the following integrals as limit of sums:

$\int_{1}^{3} (3 x^{2} + 1) d x$ [CBSE 2014]

Answer:

We have,

$\int_{a}^{b} f (x) d x = \lim_{h \to 0} \{f (a) + f (a + h) + f (a + 2 h) + . . . + f [(a + (n - 1) h)]\}$

Here, a = 1, b = 3, f(x) = 3x² + 1 and $h = \frac{3 - 1}{n} = \frac{2}{n} \Rightarrow n h = 2$

$∴ \int_{1}^{3} (3 x^{2} + 1) d x = \lim_{h \to 0} h \{f (1) + f (1 + h) + f (1 + 2 h) + . . . + f [1 + (n - 1) h]\} = \lim_{h \to 0} h \{[3 \times 1^{2} + 1] + [3 \times {(1 + h)}^{2} + 1] + [3 \times {(1 + 2 h)}^{2} + 1] + . . . + [3 \times {(1 + (n - 1) h)}^{2} + 1]\} = \lim_{h \to 0} h \{3 [1 + (1 + 2 h + h^{2}) + (1 + 4 h + 2^{2} h^{2}) + . . . + (1 + 2 (n - 1) h + {(n - 1)}^{2} h^{2})] + n\} = \lim_{h \to 0} h \{3 [n + 2 (1 + 2 + . . . + (n - 1)) h + (1^{2} + 2^{2} + . . . + {(n - 1)}^{2}) h^{2}] + n\} = \lim_{h \to 0} h [4 n + 6 \times \frac{n (n - 1)}{2} h + 3 \times \frac{(n - 1) n (2 n - 1)}{6} h^{2}]$
$= \lim_{h \to 0} [4 n h + 6 \times \frac{n h (n h - h)}{2} + 3 \times \frac{(n h - h) n h (2 n h - h)}{6}] = \lim_{h \to 0} [4 n h + 3 \times n h (n h - h) + 3 \times \frac{(n h - h) n h (2 n h - h)}{6}] = \lim_{h \to 0} [4 \times 2 + 3 \times 2 \times (2 - h) + 3 \times \frac{(2 - h) \times 2 \times (2 \times 2 - h)}{6}] = 8 + 6 \times (2 - 0) + \frac{(2 - 0) \times 2 \times (4 - 0)}{2} = 8 + 12 + 8 = 28$

Page No 20.111:

Question 34:

$\int_{1}^{3} (x^{2} + 3 x + e^{x}) d x$

Answer:

$\int_{1}^{3} (x^{2} + 3 x + e^{x}) d x \Rightarrow a = 1, b = 3, f (x) = x^{2} + 3 x + e^{x}, h = \frac{2}{n} I = \int_{1}^{3} (x^{2} + 3 x + e^{x}) d x = \lim_{h \to 0} h [f (1) + f (1 + h) + f (1 + 2 h) + . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [(4 + e) + [{(1 + h)}^{2} + 3 (1 + h) + e^{1 + h}] + [{(1 + 2 h)}^{2} + 3 (1 + 2 h) + e^{1 + 2 h}] + . . . + [{(1 + (n - 1) h)}^{2} + 3 (1 + (n - 1) h) + e^{1 + (n - 1) h}]] = \lim_{h \to 0} h [4 + e + [(n - 1) \cdot 1 + h^{2} (1 + 2^{2} + 3^{2} + . . . {(n - 1)}^{2}) + 2 h (1 + 2 + 3 + (n - 1))] + [3 (n - 1) + 3 h (1 + 2 + 3 . . . (n - 1))] + [e^{1 + h} + e^{1 + 2 h} + . . . . e^{1 + (n - 1) h}]] = \lim_{h \to 0} h [4 n + e + h^{2} \frac{(n - 1) (n - 2) (2 n - 3)}{6} + 2 h (\frac{n - 1 (n)}{2}) + 3 h (\frac{(n - 1) (n)}{2}) + e [\frac{e^{h} \cdot [e^{h (n - 1)} - 1]}{e^{h} - 1}]] = \lim_{n \to \infty} 4 n \cdot \frac{2}{n} + \frac{e \cdot 2}{n} + \frac{8}{n^{3}} \frac{(2 n^{3} . . .)}{6} + \frac{5.4}{n^{2}} (\frac{n^{2} - n}{2}) + e^{\frac{2}{n + 1}} [e^{\frac{2}{n} (n - 1)} - 1] = 8 + 0 + \frac{8}{3} + 10 + e (e^{2} - 1) = \frac{26}{3} + 12 + e^{3} - e = \frac{62}{3} + e^{3} - e$

Page No 20.115:

Question 1:

$\int_{0}^{π / 2} \sin^{2} x d x .$

Answer:

$\int_{0}^{\frac{π}{2}} \sin^{2} x d x = \int_{0}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 - \cos 2 x) d x = \frac{1}{2} {[x - \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} = \frac{1}{2} (\frac{π}{2} - 0) = \frac{π}{4}$

Page No 20.115:

Question 2:

$\int_{0}^{π / 2} \cos^{2} x d x .$

Answer:

$\int_{0}^{\frac{π}{2}} \cos^{2} x d x = \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 x}{2} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + \cos 2 x) d x = \frac{1}{2} {[x + \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} = \frac{1}{2} [\frac{π}{2} + 0] = \frac{π}{4}$

Page No 20.115:

Question 3:

$\int_{- π / 2}^{π / 2} \sin^{2} x d x .$

Answer:

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{2} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 - \cos 2 x}{2} d x = \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 - \cos 2 x) d x = \frac{1}{2} {[x - \frac{\sin 2 x}{2}]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{1}{2} (\frac{π}{2} - 0 + \frac{π}{2} - 0) = \frac{π}{2}$

Page No 20.115:

Question 4:

$\int_{- π / 2}^{π / 2} \cos^{2} x d x .$

Answer:

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + \cos 2 x}{2} d x = \frac{1}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 + \cos 2 x) d x = \frac{1}{2} {[x + \frac{\sin 2 x}{2}]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{1}{2} (\frac{π}{2} + 0 + \frac{π}{2} - 0) = \frac{π}{2}$

Page No 20.115:

Question 5:

$\int_{- π / 2}^{π / 2} \sin^{3} x d x .$

Answer:

$Let I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin^{3} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin x \sin^{2} x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin x (1 - \cos^{2} x) d x Let \cos x = t, then - \sin x d x = d t, When, x \to - \frac{π}{2}; t \to 0 and x \to \frac{π}{2}; t \to 0 I = \int_{0}^{0} (- 1 + t^{2}) d t = 0$

Page No 20.115:

Question 6:

$\int_{- π / 2}^{π / 2} x \cos^{2} x d x .$

Answer:

$We have, I = \int_{- \frac{π}{2}}^{\frac{π}{2}} x \cos^{2} x d x Let f (x) = x \cos^{2} x \Rightarrow f (- x) = (- x) \cos^{2} (- x) = - x \cos^{2} x ∴ f (- x) = - f (x) i . e ., f (x) is odd function We know that \int_{- a}^{a} f (x) d x = 0, if f (x) is odd function . ∴ I = \int_{- \frac{π}{2}}^{\frac{π}{2}} x \cos^{2} x d x = 0$

Page No 20.115:

Question 7:

$\int_{0}^{π / 4} \tan^{2} x d x .$

Answer:

$\int_{0}^{\frac{π}{4}} \tan^{2} x d x = \int_{0}^{\frac{π}{4}} (s e c^{2} x - 1) d x = {[\tan x - x]}_{0}^{\frac{π}{4}} = 1 - \frac{π}{4} - 0 = 1 - \frac{π}{4}$ 4430. →

Page No 20.115:

Question 8:

$\int_{0}^{1} \frac{1}{x^{2} + 1} d x$ .

Answer:

$\int_{0}^{1} \frac{1}{1 + x^{2}} d x = {[\tan^{- 1} x]}_{0}^{1} = \frac{π}{4} - 0 = \frac{π}{4}$ ode is 4430. →

Page No 20.115:

Question 9:

$\int_{- 2}^{1} \frac{|x|}{x} d x .$

Answer:

$Let, I = \int_{- 2}^{1} \frac{|x|}{x} d x We have, |x| = \{\begin{cases} x 0 \leq x \leq 1 \\ - x - 2 \leq x < 0 \end{cases} ∴ \frac{|x|}{x} = \{\begin{cases} 1 0 \leq x \leq 1 \\ - 1 - 2 \leq x < 0 \end{cases} Therefore, I = \int_{- 2}^{0} - 1 d x + \int_{0}^{1} 1 d x = - {[x]}_{- 2}^{0} + {[x]}_{0}^{1} = 0 - 2 + 1 - 0 = - 1$

Page No 20.115:

Question 10:

$\int_{0}^{\infty} e^{- x} d x .$

Answer:

$\int_{0}^{\infty} e^{- x} d x = - {[e^{- x}]}_{0}^{\infty} = - (0 - 1) = 0 + 1 = 1$ 4430. →

Page No 20.115:

Question 11:

$\int_{0}^{4} \frac{1}{\sqrt{16 - x^{2}}} d x .$

Answer:

$\int_{0}^{4} \frac{1}{\sqrt{16 - x^{2}}} d x = \int_{0}^{4} \frac{1}{\sqrt{4^{2} - x^{2}}} d x = {[\sin^{- 1} \frac{x}{4}]}_{0}^{4} = (\frac{π}{2} - 0) = \frac{π}{2}$ s 4430. →

Page No 20.115:

Question 12:

$\int_{0}^{3} \frac{1}{x^{2} + 9} d x .$

Answer:

$\int_{0}^{3} \frac{1}{x^{2} + 9} d x = \int_{0}^{3} \frac{1}{x^{2} + 3^{2}} d x = \frac{1}{3} {[\tan^{- 1} \frac{x}{3}]}_{0}^{3} = \frac{1}{3} (\tan^{- 1} 1 - \tan^{- 1} 0) = \frac{1}{3} (\frac{π}{4} - 0) = \frac{π}{12}$

Page No 20.115:

Question 13:

$\int_{0}^{π / 2} \sqrt{1 - \cos 2 x} d x .$

Answer:

$\int_{0}^{\frac{π}{2}} \sqrt{1 - \cos 2 x} d x = \int_{0}^{\frac{π}{2}} \sqrt{2 \sin^{2} x} d x = \int_{0}^{\frac{π}{2}} \sqrt{2} \sin x d x = - \sqrt{2} {[\cos x]}_{0}^{\frac{π}{2}} = - (0 - \sqrt{2}) = \sqrt{2}$

Page No 20.115:

Question 14:

$\int_{0}^{π / 2} \log \tan x d x .$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \log \tan x d x . . . (i) = \int_{0}^{\frac{π}{2}} \log \tan (\frac{π}{2} - x) d x [Using, \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] = \int_{0}^{\frac{π}{2}} \log c o t x d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{\frac{π}{2}} \log \tan x d x + \int_{0}^{\frac{π}{2}} \log c o t x d x = \int_{0}^{\frac{π}{2}} \log (\tan x \times c o t x) d x = \int_{0}^{\frac{π}{2}} \log 1 d x = 0 Hence, I = 0$

Page No 20.115:

Question 15:

$\int_{0}^{π / 2} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) d x .$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) d x . . . (i) = \int_{0}^{\frac{π}{2}} l og [\frac{3 + 5 \cos (\frac{π}{2} - x)}{3 + 5 \sin (\frac{π}{2} - x)}] d x = \int_{0}^{\frac{π}{2}} l og (\frac{3 + 5 \sin x}{3 + 5 \cos x}) d x . . . (ii) Adding (i) and (i i) 2 I = \int_{0}^{\frac{π}{2}} [\log (\frac{3 + 5 \cos x}{3 + 5 \sin x}) + l og (\frac{3 + 5 \sin x}{3 + 5 \cos x})] d x = \int_{0}^{\frac{π}{2}} \log (\frac{3 + 5 \cos x}{3 + 5 \sin x} \times \frac{3 + 5 \sin x}{3 + 5 \cos x}) d x = \int_{0}^{\frac{π}{2}} \log 1 d x = 0 Hence I = 0$

Page No 20.115:

Question 16:

$\int_{0}^{π / 2} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x, n \in N .$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (\frac{π}{2} - x)}{\sin^{n} (\frac{π}{2} - x) + \cos^{n} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\cos^{n} x + \sin^{n} x} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} [\frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} + \frac{\cos^{n} x}{\sin^{n} x + \cos^{n} x}] d x = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x + \cos^{n} x}{\sin^{n} x + \cos^{n} x} d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 20.115:

Question 17:

$\int_{0}^{π} \cos^{5} x d x .$

Answer:

$Let I = \int_{0}^{π} \cos^{5} x d x = \int_{0}^{π} \cos x {(\cos^{2} x)}^{2} d x = \int_{0}^{π} \cos x {(1 - \sin^{2} x)}^{2} d x Let \sin x = t, then \cos x d x = d t When, x \to 0; t \to 0 and x \to π; t \to 0 Therefore, I = \int_{0}^{0} {(1 - t^{2})}^{2} d t = 0$

Page No 20.115:

Question 18:

$\int_{- π / 2}^{π / 2} \log (\frac{a - \sin θ}{a + \sin θ}) d θ$

Answer:

$Let, I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \log (\frac{a - \sin θ}{a + \sin θ}) d θ Here, f (θ) = \log (\frac{a - \sin θ}{a + \sin θ}) Consider, f (- θ) = \log [\frac{a - \sin (- θ)}{a + \sin (- θ)}] = - \log (\frac{a - \sin θ}{a + \sin θ}) = - f (θ) i . e ., f (θ) is odd function . Therefore, I = 0$

Page No 20.115:

Question 19:

$\int_{- 1}^{1} x |x| d x .$

Answer:

$|x| = \{\begin{cases} - x, - 1 < x < 0 \\ x, 0 < x < 1 \end{cases} ∴ x |x| = \{\begin{cases} - x^{2}, - 1 < x < 0 \\ x^{2}, 0 < x < 1 \end{cases} Now, \int_{- 1}^{1} x |x| d x = \int_{- 1}^{0} - x^{2} d x + \int_{0}^{1} x^{2} d x = - \int_{- 1}^{0} x^{2} d x + \int_{0}^{1} x^{2} d x = - {[\frac{x^{3}}{3}]}_{- 1}^{0} + {[\frac{x^{3}}{3}]}_{0}^{1} = - (0 + \frac{1}{3}) + (\frac{1}{3} - 0) = 0 - \frac{1}{3} + \frac{1}{3} - 0 = 0$

Page No 20.115:

Question 20:

$\int_{a}^{b} \frac{f (x)}{f (x) + f (a + b - x)} d x .$

Answer:

$Let I = \int_{a}^{b} \frac{f (x)}{f (x) + f (a + b - x)} d x . . . (i) = \int_{a}^{b} \frac{f (a + b - x)}{f (a + b - x) + f (a + b - a - b + x)} d x = \int_{a}^{b} \frac{f (a + b - x)}{f (a + b - x) + f (x)} d x ∴ I = \int_{a}^{b} \frac{f (a + b - x)}{f (x) + f (a + b - x)} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{a}^{b} [\frac{f (x)}{f (x) + f (a + b - x)} + \frac{f (a + b - x)}{f (x) + f (a + b - x)}] d x = \int_{a}^{b} \frac{f (x) + f (a + b - x)}{f (x) + f (a + b - x)} dx = {[x]}_{a}^{b} = b - a Hence, I = \frac{b - a}{2}$

Page No 20.115:

Question 21:

$\int_{0}^{1} \frac{1}{1 + x^{2}} d x$

Answer:

$\int_{0}^{1} \frac{1}{1 + x^{2}} d x = {[\tan^{- 1} x]}_{0}^{1} = \tan^{- 1} 1 - \tan^{- 1} 0 = \frac{π}{4} - 0 = \frac{π}{4}$

Page No 20.115:

Question 22:

Evaluate each of the following integrals:

$\int_{0}^{\frac{π}{4}} \tan x d x$

Answer:

$\int_{0}^{\frac{π}{4}} \tan x d x = {logsec x}_{0}^{\frac{π}{4}} = logsec \frac{π}{4} - logsec 0 = \log \sqrt{2} - \log 1 = \log 2^{\frac{1}{2}} - 0 = \frac{1}{2} \log 2$

Page No 20.115:

Question 23:

$\int_{2}^{3} \frac{1}{x} d x$

Answer:

$\int_{2}^{3} \frac{1}{x} d x = {[\log_{e} x]}_{2}^{3} = \log_{e} 3 - \log_{e} 2 = \log_{e} (\frac{3}{2})$ →

Page No 20.115:

Question 24:

$\int_{0}^{2} \sqrt{4 - x^{2}} d x$

Answer:

$\int_{0}^{2} \sqrt{4 - x^{2}} d x = \int_{0}^{2} \sqrt{2^{2} - x^{2}} d x = {[\frac{x}{2} \sqrt{4 - x^{2}} + \frac{1}{2} \times 2^{2} \sin^{- 1} \frac{x}{2}]}_{0}^{2} = {[\frac{x}{2} \sqrt{4 - x^{2}}]}_{0}^{2} + 2 {[\sin^{- 1} \frac{x}{2}]}_{0}^{2} = 0 + 2 (\frac{π}{2} - 0) = π$

Page No 20.115:

Question 25:

$\int_{0}^{1} \frac{2 x}{1 + x^{2}} d x$

Answer:

$We have, I = \int_{0}^{1} \frac{2 x}{1 + x^{2}} d x Putting 1 + x^{2} = t \Rightarrow 2 x d x = d t When x \to 0; t \to 1 And x \to 1; t \to 2 ∴ I = \int_{1}^{2} \frac{d t}{t} = {[\log_{e} |t|]}_{1}^{2} = \log_{e} 2 - \log_{e} 1 = \log_{e} 2 - 0 = \log_{e} 2$ 4430. →

Page No 20.115:

Question 26:

Evaluate each of the following integrals:

$\int_{0}^{1} x e^{x^{2}} d x$ [CBSE 2014]

Answer:

$I = \int_{0}^{1} x e^{x^{2}} d x = \frac{1}{2} \int_{0}^{1} e^{x^{2}} 2 x d x$

Put $x^{2} = z$

$\Rightarrow 2 x d x = d z$

When $x \to 0, z \to 0$

When $x \to 1, z \to 1$

$∴ I = \frac{1}{2} \int_{0}^{1} e^{z} d z = \frac{1}{2} \times {e^{z}}_{0}^{1} = \frac{1}{2} (e - e^{0}) = \frac{1}{2} (e - 1)$

Page No 20.115:

Question 27:

Evaluate each of the following integrals:

$\int_{0}^{\frac{π}{4}} \sin 2 x d x$ [CBSE 2014]

Answer:

$\int_{0}^{\frac{π}{4}} \sin 2 x d x = {\frac{- \cos 2 x}{2}}_{0}^{\frac{π}{4}} = - \frac{1}{2} (\cos \frac{π}{2} - \cos 0) = - \frac{1}{2} \times (0 - 1) = \frac{1}{2}$

Page No 20.115:

Question 28:

Evaluate each of the following integrals:

$\int_{e}^{e^{2}} \frac{1}{x \log x} d x$ [CBSE 2014]

Answer:

$\int_{e}^{e^{2}} \frac{1}{x \log x} d x = \int_{e}^{e^{2}} \frac{\frac{1}{x}}{\log x} d x = {\log (\log x)}_{e}^{e^{2}} [\int \frac{f' (x)}{f (x)} d x = \log f (x) + C] = \log (\log e^{2}) - \log (\log e) = \log (2 \log e) - \log (\log e) = \log 2 - \log 1 (\log e = 1) = \log 2 - 0 = \log 2$

Page No 20.115:

Question 29:

Evaluate each of the following integrals:

$\int_{0}^{\frac{π}{2}} e^{x} (\sin x - \cos x) d x$ [CBSE 2014]

Answer:

Disclaimer: The solution has been provided by taking the lower limit of integral as 0.

$\int_{0}^{\frac{π}{2}} e^{x} (\sin x - \cos x) d x = - \int_{0}^{\frac{π}{2}} e^{x} [\cos x + (- \sin x)] d x = - {e^{x} \cos x}_{0}^{\frac{π}{2}} \{\int e^{x} [f (x) + f' (x)] d x = e^{x} f (x) + C\} = - (e^{\frac{π}{2}} \cos \frac{π}{2} - e^{0} \cos 0) = - (e^{\frac{π}{2}} \times 0 - 1 \times 1) = - (0 - 1) = 1$

Page No 20.115:

Question 30:

Solve each of the following integrals:

$\int_{2}^{4} \frac{x}{x^{2} + 1} d x$ [CBSE 2014]

Answer:

$\int_{2}^{4} \frac{x}{x^{2} + 1} d x = \frac{1}{2} \int_{2}^{4} \frac{2 x}{x^{2} + 1} d x = \frac{1}{2} \times {\log (x^{2} + 1)}_{2}^{4} [\int \frac{f' (x)}{f (x)} d x = \log f (x) + C] = \frac{1}{2} (\log 17 - \log 5) = \frac{1}{2} \log (\frac{17}{5}) (\log a - \log b = \log \frac{a}{b})$

Page No 20.116:

Question 31:

If $\int_{0}^{1} (3 x^{2} + 2 x + k) d x = 0,$ find the value of k.

Answer:

$We have, \int_{0}^{1} (3 x^{2} + 2 x + k) d x = 0 \Rightarrow {[x^{3} + x^{2} + k x]}_{0}^{1} = 0 \Rightarrow 1 + 1 + k - 0 = 0 \Rightarrow k = - 2$ . →

Page No 20.116:

Question 32:

If $\int_{0}^{a} 3 x^{2} d x = 8,$ write the value of a.

Answer:

$We have, \int_{0}^{a} 3 x^{2} d x = 8 \Rightarrow {[3 \frac{x^{3}}{3}]}_{0}^{a} = 8 \Rightarrow {[x^{3}]}_{0}^{a} = 8 \Rightarrow a^{3} - 0 = 8 \Rightarrow a = \sqrt[3]{8} = 2$ 0. →

Page No 20.116:

Question 33:

If $f (x) = \int_{0}^{x} t \sin t d t$ , the write the value of $f' (x)$ . [CBSE 2014]

Answer:

$f (x) = \int_{0}^{x} t \sin t d t \Rightarrow f (x) = {t (- \cos t)}_{0}^{x} - \int_{0}^{x} \frac{d}{d t} (t) \times (- \cos t) d t \Rightarrow f (x) = - (x \cos x - 0) + \int_{0}^{x} \cos t d t \Rightarrow f (x) = - x \cos x + {\sin t}_{0}^{x}$
$\Rightarrow f (x) = - x \cos x + (\sin x - 0) \Rightarrow f (x) = - x \cos x + \sin x$

Differentiating both sides with respect to x, we get

$f' (x) = - [x \times (- \sin x) + \cos x \times 1] + \cos x \Rightarrow f' (x) = - (- x \sin x) - \cos x + \cos x \Rightarrow f' (x) = x \sin x$

Thus, the value of $f' (x)$ is x sinx.

Page No 20.116:

Question 34:

If $\int_{0}^{a} \frac{1}{4 + x^{2}} d x = \frac{π}{8}$ , find the value of a. [CBSE 2014]

Answer:

$\int_{0}^{a} \frac{1}{4 + x^{2}} d x = \frac{π}{8} \Rightarrow {\frac{1}{2} \tan^{- 1} \frac{x}{2}}_{0}^{a} = \frac{π}{8} [\int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C] \Rightarrow \frac{1}{2} (\tan^{- 1} \frac{a}{2} - \tan^{- 1} 0) = \frac{π}{8} \Rightarrow \tan^{- 1} \frac{a}{2} - 0 = \frac{π}{4}$
$\Rightarrow \tan^{- 1} \frac{a}{2} = \frac{π}{4} \Rightarrow \frac{a}{2} = \tan \frac{π}{4} = 1 \Rightarrow a = 2$

Thus, the value of a is 2.

Page No 20.116:

Question 35:

Write the coefficient a, b, c of which the value of the integral $\int_{- 3}^{3} (a x^{2} + b x + c) d x$ is independent.

Answer:

$\int_{- 3}^{3} (a x^{2} + b x + c) d x = {[a \frac{x^{3}}{3} + b \frac{x^{2}}{2} + c x]}_{- 3}^{3} = 9 a + \frac{9}{2} b + 3 c + 9 a - \frac{9}{2} b + 3 c = 18 a + 6 c$

Hence, the given integral is independent of b

Page No 20.116:

Question 36:

Evaluate : $\int_{2}^{3} 3^{x} d x .$

Answer:

$I = \int_{2}^{3} 3^{x} = {\frac{3^{x}}{\log 3}}_{2}^{3} + C (Use : \int a^{x} = \frac{a^{x}}{\log a} + C) = \frac{3^{3}}{\log 3} - \frac{3^{2}}{\log 3} + C$

$= \frac{1}{\log 3} (3^{3} - 3^{2}) + C = \frac{1}{\log 3} (27 - 9) + C = \frac{1}{\log 3} (18) + C$

Page No 20.116:

Question 37:

$\int_{0}^{2} [x] d x .$

Answer:

$We have, I = \int_{0}^{2} [x] d x We know that, [x] = \{\begin{cases} 0, 0 < x < 1 \\ 1, 1 < x < 2 \end{cases} ∴ I = \int_{0}^{2} [x] d x = \int_{0}^{1} [x] d x + \int_{1}^{2} [x] d x = \int_{0}^{1} (0) d x + \int_{1}^{2} (1) d x = 0 + {[x]}_{1}^{2} = 2 - 1 = 1$

Page No 20.116:

Question 38:

$\int_{0}^{15} [x] d x .$

Answer:

$We have, I = \int_{0}^{1.5} [x] d x = \int_{0}^{1} [x] d x + \int_{1}^{1.5} [x] d x = \int_{0}^{1} (0) d x + \int_{1}^{1.5} (1) d x [∵ [x] = \{\begin{cases} 0 0 \leq x < 1 \\ 1 1 \leq x < 1.5 \end{cases}] = 0 + {[x]}_{1}^{1.5} = 1.5 - 1 = 0.5 = \frac{1}{2}$

Page No 20.116:

Question 39:

$\int_{0}^{1} \{x\} d x,$ where {x} denotes the fractional part of x.

Answer:

$We have, I = \int_{0}^{1} \{x\} d x We know \{x\} = x, 0 < x < 1 ∴ I = \int_{0}^{1} x d x = {[\frac{x^{2}}{2}]}_{0}^{1} = \frac{1}{2} - \frac{0}{2} = \frac{1}{2}$
s 4430. →

Page No 20.116:

Question 40:

$\int_{0}^{1} e^{\{x\}} d x .$

Answer:

$We have, I = \int_{0}^{1} e^{\{x\}} d x We know that, \{x\} = x, when 0 < x < 1 ∴ I = \int_{0}^{1} e^{x} d x = {[e^{x}]}_{0}^{1} = e^{1} - e^{0} = e - 1$

Page No 20.116:

Question 41:

$\int_{0}^{2} x [x] d x .$

Answer:

$We have, I = \int_{0}^{2} x [x] d x We know that, x [x] = \{\begin{cases} x \times 0, 0 < x < 1 \\ x \times 1, 1 < x < 2 \end{cases} i . e ., x [x] = \{\begin{cases} 0, 0 < x < 1 \\ x, 1 < x < 2 \end{cases} ∴ I = \int_{0}^{2} x [x] d x = \int_{0}^{1} x [x] d x + \int_{1}^{2} x [x] d x = \int_{0}^{1} (0) d x + \int_{1}^{2} (x) d x = 0 + {[\frac{x^{2}}{2}]}_{1}^{2} = \frac{2^{2}}{2} - \frac{1^{2}}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2}$

Page No 20.116:

Question 42:

$\int_{0}^{1} 2^{x - [x]} d x$

Answer:

$We have, I = \int_{0}^{1} 2^{x - [x]} d x = \int_{0}^{1} 2^{x - 0} d x (∵ [x] = 0 where, 0 < x < 1) = \int_{0}^{1} 2^{x} d x = {[\frac{2^{x}}{\log_{e} 2}]}_{0}^{1} = \frac{2^{1}}{\log_{e} 2} - \frac{2^{0}}{\log_{e} 2} = \frac{2}{\log_{e} 2} - \frac{1}{\log_{e} 2} = \frac{1}{\log_{e} 2}$

Page No 20.116:

Question 43:

$\int_{1}^{2} \log_{e} [x] d x .$

Answer:

$We have, I = \int_{1}^{2} \log_{e} [x] d x We know that, [x] = 1, when 1 < x < 2 ∴ I = \int_{1}^{2} \log_{e} 1 d x I = \int_{1}^{2} (0) d x = 0$

Page No 20.116:

Question 44:

$\int_{0}^{\sqrt{2}} [x^{2}] d x .$

Answer:

$We have, I = \int_{0}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x = \int_{0}^{1} (0) d x + \int_{1}^{\sqrt{2}} (1) d x (∵ [x^{2}] = \{\begin{cases} 0 0 < x < 1 \\ 1 1 < x < \sqrt{2} \end{cases}) = 0 + {[x]}_{1}^{\sqrt{2}} = \sqrt{2} - 1$

Page No 20.116:

Question 45:

If $[\cdot] and \{\cdot\}$ denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
$\int_{0}^{π / 4} \sin \{x\} d x$

Answer:

$We have, I = \int_{0}^{π / 4} \sin \{x\} d x We know that, \{x\} = x, when 0 < x < \frac{π}{4} (As π = 3.14 \Rightarrow \frac{π}{4} = 0.785 < 1) ∴ I = \int_{0}^{π / 4} \sin x d x = {[- \cos x]}_{0}^{\frac{π}{4}} = - (\cos \frac{π}{4} - \cos 0) = \cos 0 - \cos \frac{π}{4} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}}$

Page No 20.117:

Question 1:

$\int_{0}^{1} \sqrt{x (1 - x)} d x$ equals

(a) π/2
(b) π/4
(c) π/6
(d) π/8

Answer:

(d) $π$ /8

$Let, I = \int_{0}^{1} \sqrt{x (1 - x)} d x = \int_{0}^{1} \sqrt{x - x^{2}} d x = \int_{0}^{1} \sqrt{\frac{1}{4} - (x^{2} - x + \frac{1}{4})} d x = \int_{0}^{1} \sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{1}{2})}^{2}} d x = {[\frac{(x - \frac{1}{2})}{2} \sqrt{x - x^{2}} + \frac{1}{2} \times \frac{1}{4} \sin^{- 1} (2 x - 1)]}_{0}^{1} = \frac{1}{8} {[\sin^{- 1} (1) - \sin^{- 1} (- 1)]}_{0}^{1} = \frac{1}{8} [\frac{π}{2} + \frac{π}{2}] = \frac{π}{8}$

Page No 20.117:

Question 2:

$\int_{0}^{π} \frac{1}{1 + \sin x} d x$ equals

(a) 0
(b) 1/2
(c) 2
(d) 3/2

Answer:

(c) 2

$\int_{0}^{π} \frac{1}{1 + \sin x} d x = \int_{0}^{π} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} d x = \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} d x = \int_{0}^{π} \frac{1 - \sin x}{\cos^{2} x} d x = \int_{0}^{π} (s e c^{2} x - s e c x \tan x) d x = {[\tan x - s e c x]}_{0}^{π} = 0 + 1 - 0 + 1 = 2$

Page No 20.117:

Question 3:

The value of $\int_{0}^{π} \frac{x \tan x}{\sec x + \cos x} d x$ is
(a) $\frac{π^{2}}{4}$

(b) $\frac{π^{2}}{2}$

(c) $\frac{3 π^{2}}{2}$

(d) $\frac{π^{2}}{3}$

Answer:

$(a) \frac{π^{2}}{4}$ π24

$Let I = \int_{0}^{π} \frac{x \tan x}{s e c x + \cos x} d x (i) = \int_{0}^{π} \frac{(π - x) \tan (π - x)}{s e c (π - x) + \cos (π - x)} d x = \int_{0}^{π} \frac{(π - x) tanx}{s e c x + \cos x} d x (i i) Adding (i) and (i i) 2 I = \int_{0}^{π} \frac{x \tan x}{s e c x + \cos x} + \frac{(π - x) tanx}{s e c x + \cos x} d x = \int_{0}^{π} \frac{π \tan x}{s e c x + \cos x} d x = π \int_{0}^{π} \frac{sinx}{1 + \cos^{2} x} dx = - π {[\tan^{- 1} cosx]}_{0}^{π} = - π (- \frac{π}{4} - \frac{π}{4}) = \frac{π^{2}}{2} Hence I = \frac{π^{2}}{4}$ $We have, I = \int_{0}^{π} \frac{x \tan x}{\sec x + \cos x} d x . . . . . (1) = \int_{0}^{π} \frac{(π - x) \tan (π - x)}{\sec (π - x) + \cos (π - x)} d x = \int_{0}^{π} \frac{(π - x) tanx}{\sec x + \cos x} d x . . . . . (2) Adding (1) and (2), we get 2 I = \int_{0}^{π} [\frac{x \tan x}{\sec x + \cos x} + \frac{(π - x) tanx}{\sec x + \cos x}] d x \Rightarrow I = \frac{1}{2} \int_{0}^{π} \frac{π \tan x}{\sec x + \cos x} d x = \frac{π}{2} \int_{0}^{π} \frac{sinx}{1 + \cos^{2} x} dx Putting \cos x = t \Rightarrow - \sin x d x = d t \Rightarrow \sin x d x = - d t When x \to 0; t \to 1 and x \to π; t \to - 1 \Rightarrow I = \frac{π}{2} \int_{1}^{- 1} \frac{- d t}{1 + t^{2}} = \frac{π}{2} \int_{- 1}^{1} \frac{d t}{1 + t^{2}} = \frac{π}{2} {[\tan^{- 1} t]}_{- 1}^{1} = \frac{π}{2} [\tan^{- 1} (1) - \tan^{- 1} (- 1)] = \frac{π}{2} [\frac{π}{4} - (- \frac{π}{4})] = \frac{π}{2} \times \frac{π}{2} = \frac{π^{2}}{4} Hence I = \frac{π^{2}}{4}$

Page No 20.117:

Question 4:

The value of $\int_{0}^{2 π} \sqrt{1 + \sin \frac{x}{2}} d x$ is
(a) 0
(b) 2
(c) 8
(d) 4

Answer:

(c) 8

$\int_{0}^{2 π} \sqrt{1 + \sin \frac{x}{2}} d x = \int_{0}^{2 π} \sqrt{\sin^{2} \frac{x}{4} + \cos^{2} \frac{x}{4} + 2 \sin \frac{x}{4} \cos \frac{x}{4}} d x = \int_{0}^{2 π} (\sin \frac{x}{4} + \cos \frac{x}{4}) d x = {[\frac{- \cos \frac{x}{4}}{\frac{1}{4}} + \frac{\sin \frac{x}{4}}{\frac{1}{4}}]}_{0}^{2 π} = 4 {[\sin \frac{x}{4} - \cos \frac{x}{4}]}_{0}^{2 π} = 4 [\sin \frac{2 π}{4} - \cos \frac{2 π}{4} - \sin 0 + \cos 0] = 4 [\sin \frac{π}{2} - \cos \frac{π}{2} - 0 + 1] = 4 [1 - 0 - 0 + 1] = 4 \times 2 = 8$

Page No 20.117:

Question 5:

The value of the integral $\int_{0}^{π / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x$ is
(a) 0
(b) π/2
(c) π/4
(d) none of these

Answer:

(c) π/4

$Let I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\cos (\frac{π}{2} - x)}}{\sqrt{\cos (\frac{π}{2} - x)} + \sqrt{\sin (\frac{π}{2} - x)}} d x = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} d x = \int_{0}^{\frac{π}{2}} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} [\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} + \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}] d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 20.117:

Question 6:

$\int_{0}^{\infty} \frac{1}{1 + e^{x}} d x$ equals

(a) log 2 − 1
(b) log 2
(c) log 4 − 1
(d) − log 2

Answer:

(b) log 2

$We have, I = \int_{0}^{\infty} \frac{1}{1 + e^{x}} d x Putting e^{x} = t \Rightarrow e^{x} d x = d t \Rightarrow d x = \frac{d t}{t} When x \to 0; t \to 1 and x \to \infty; t \to \infty ∴ I = \int_{1}^{\infty} \frac{1}{t (1 + t)} d t = \int_{1}^{\infty} \frac{1}{t + t^{2}} d t = \int_{1}^{\infty} \frac{1}{{(t + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d t$

$= \frac{1}{2 \times \frac{1}{2}} {[\log |\frac{t + \frac{1}{2} - \frac{1}{2}}{t + \frac{1}{2} + \frac{1}{2}}|]}_{1}^{\infty} = {[\log |\frac{t}{t + 1}|]}_{1}^{\infty} = {[\log |\frac{\frac{t}{t}}{\frac{t}{t} + \frac{1}{t}}|]}_{1}^{\infty} = {[\log |\frac{1}{1 + \frac{1}{t}}|]}_{1}^{\infty} = \log \frac{1}{1 + 0} - \log \frac{1}{1 + 1} = \log (1) - \log (\frac{1}{2}) = 0 - (- \log 2) = \log 2$

Page No 20.117:

Question 7:

$\int_{0}^{π^{2} / 4} \frac{\sin \sqrt{x}}{\sqrt{x}} d x$ equals
(a) 2
(b) 1
(c) π/4
(d) π²/8

Answer:

(a) 2

$\int_{0}^{\frac{π^{2}}{4}} \frac{\sin \sqrt{x}}{\sqrt{x}} d x Let \sqrt{x} = t, then \frac{1}{2 \sqrt{x}} d x = d t When x = 0, t = 0, x = \frac{π^{2}}{4}, t = \frac{π}{2} Therefore the integral becomes \int_{0}^{\frac{π}{2}} 2 sint dt = - 2 {[cost]}_{0}^{\frac{π}{2}} = 2$

Page No 20.117:

Question 8:

$\int_{0}^{π / 2} \frac{\cos x}{(2 + \sin x) (1 + \sin x)} d x$ equals

(a) $\log (\frac{2}{3})$

(b) $\log (\frac{3}{2})$

(c) $\log (\frac{3}{4})$

(d) $\log (\frac{4}{3})$

Answer:

(d) $\log (\frac{4}{3})$

$Let, I = \int_{0}^{\frac{π}{2}} \frac{\cos x}{(2 + \sin x) (1 + \sin x)} d x Let \sin x, then \cos x d x = d t When x = 0, t = 0, x = \frac{π}{2}, t = 1 Therefore the integral becomes I = \int_{0}^{1} \frac{dt}{(2 + t) (1 + t)} = \int_{0}^{1} [\frac{- 1}{2 + t} + \frac{1}{1 + t}] dt = {[- \log (2 + t) + \log (1 + t)]}_{0}^{1} = {[\log (1 + t) - \log (2 + t)]}_{0}^{1} = \log 2 - \log 3 - \log 1 + \log 2 = \log \frac{4}{3}$

Page No 20.117:

Question 9:

$\int_{0}^{π / 2} \frac{1}{2 + \cos x} d x$ equals

(a) $\frac{1}{3} \tan^{- 1} (\frac{1}{\sqrt{3}})$

(b) $\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}})$

(c) $\sqrt{3} \tan^{- 1} (\sqrt{3})$

(d) $2 \sqrt{3} \tan^{- 1} \sqrt{3}$

Answer:

$(b) \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}})$

$We have, I = \int_{0}^{\frac{π}{2}} \frac{1}{2 + \cos x} d x = \int_{0}^{\frac{π}{2}} \frac{1}{2 + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{2 + 2 \tan^{2} \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} \frac{x}{2}}{3 + \tan^{2} \frac{x}{2}} d x$

$Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t When, x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1 ∴ I = \int_{0}^{1} \frac{2}{3 + t^{2}} d t = 2 \int_{0}^{1} \frac{1}{{(\sqrt{3})}^{2} + t^{2}} d t = \frac{2}{\sqrt{3}} {[\tan^{- 1} \frac{t}{\sqrt{3}}]}_{0}^{1} = \frac{2}{\sqrt{3}} [\tan^{- 1} \frac{1}{\sqrt{3}} - \tan^{- 1} \frac{0}{\sqrt{3}}] = \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}})$
3√tan−1(13√)

Page No 20.117:

Question 10:

$\int_{0}^{π} \sqrt{\frac{1 - x}{1 + x}} d x =$

(a) $\frac{π}{2}$

(b) $\frac{π}{2} - 1$

(c) $\frac{π}{2} + 1$

(d) π + 1

Answer:

Disclaimer: None of the given option is correct.

$We have, I = \int_{0}^{π} \sqrt{\frac{1 - x}{1 + x}} d x = \int_{0}^{π} [\sqrt{\frac{1 - x}{1 + x}} \times \frac{\sqrt{1 - x}}{\sqrt{1 - x}}] d x = \int_{0}^{π} \frac{1 - x}{\sqrt{1 - x^{2}}} d x = \int_{0}^{π} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{0}^{π} \frac{x}{\sqrt{1 - x^{2}}} d x Putting 1 - x^{2} = t \Rightarrow - 2 x d x = d t \Rightarrow x d x = \frac{- d t}{2} When x \to 0; t \to 1 and x \to π; t \to 1 - π^{2} ∴ I = \int_{0}^{π} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{1}^{(1 - π^{2})} \frac{- d t}{2 \sqrt{t}} = {[\sin^{- 1} x]}_{0}^{π} + \frac{2}{2} {[\sqrt{t}]}_{1}^{(1 - π^{2})} = [0 - 0] + [\sqrt{1 - π^{2}} - \sqrt{1}] = \sqrt{1 - π^{2}} - 1$

Page No 20.117:

Question 11:

$\int_{0}^{π} \frac{1}{a + b \cos x} d x =$

(a) $\frac{π}{\sqrt{a^{2} - b^{2}}}$

(b) $\frac{π}{a b}$

(c) $\frac{π}{a^{2} + b^{2}}$

(d) (a + b) π

Answer:

$(a) \frac{π}{\sqrt{a^{2} - b^{2}}} We have, I = \int_{0}^{π} \frac{1}{a + b \cos x} d x = \int_{0}^{π} \frac{1}{a + b \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x$

$= \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{a (1 + \tan^{2} \frac{x}{2}) + b (1 - \tan^{2} \frac{x}{2})} d x = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{(a + b) + (a - b) \tan^{2} \frac{x}{2}} d x = \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{(a + b) + (a - b) \tan^{2} \frac{x}{2}} d x$

$Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t When x \to 0; t \to 0 and x \to π; t \to \infty ∴ I = \int_{0}^{\infty} \frac{2 d t}{(a + b) + (a - b) t^{2}} = \frac{2}{a - b} \int_{0}^{\infty} \frac{1}{(\frac{a + b}{a - b}) + t^{2}} d t$

$= \frac{2}{(a - b)} \int_{0}^{\infty} \frac{1}{{(\sqrt{\frac{a + b}{a - b}})}^{2} + t^{2}} d t = \frac{2}{(a - b)} \times \sqrt{\frac{a - b}{a + b}} {[\tan^{- 1} \frac{t}{\sqrt{\frac{a + b}{a - b}}}]}_{0}^{\infty} = \frac{2}{\sqrt{a^{2} - b^{2}}} [\frac{π}{2} - 0] = \frac{2}{\sqrt{a^{2} - b^{2}}} [\frac{π}{2}] = \frac{π}{\sqrt{a^{2} - b^{2}}}$

Page No 20.118:

Question 12:

$\int_{π / 6}^{π / 3} \frac{1}{1 + \sqrt{\cot} x} d x$ is

(a) π/3
(b) π/6
(c) π/12
(d) π/2

Answer:

$(c) \frac{π}{12} Let, I = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{c o t x}} d x . . . (i) = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{c o t (\frac{π}{3} + \frac{π}{6} - x)}} d x [Using \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x] = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \sqrt{\tan x}} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{\frac{π}{6}}^{\frac{π}{3}} [\frac{1}{1 + \sqrt{c o t x}} + \frac{1}{1 + \sqrt{\tan x}}] d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \frac{2 + \sqrt{c o t x} + \sqrt{\tan x}}{(1 + \sqrt{c o t x}) (1 + \sqrt{\tan x})} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} [\frac{2 + \sqrt{c o t x} + \sqrt{\tan x}}{2 + \sqrt{c o t x} + \sqrt{\tan x}}] d x = \int_{\frac{π}{6}}^{\frac{π}{3}} d x = {[x]}_{\frac{π}{6}}^{\frac{π}{3}} = \frac{π}{3} - \frac{π}{6} = \frac{π}{6} Hence, I = \frac{π}{12}$

Page No 20.118:

Question 13:

Given that $\int_{0}^{\infty} \frac{x^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2}) (x^{2} + c^{2})} d x = \frac{π}{2 (a + b) (b + c) (c + a)},$ the value of $\int_{0}^{\infty} \frac{d x}{(x^{2} + 4) (x^{2} + 9)},$ is

(a) $\frac{π}{60}$

(b) $\frac{π}{20}$

(c) $\frac{π}{40}$

(d) $\frac{π}{80}$

Answer:

$(b) \frac{π}{60} \int_{0}^{\infty} \frac{1}{(x^{2} + 4) (x^{2} + 9)} d x = \frac{1}{5} \int_{0}^{\infty} \frac{1}{(x^{2} + 4)} - \frac{1}{(x^{2} + 9)} d x = \frac{1}{5} {[\frac{1}{2} \tan^{- 1} \frac{x}{2} - \frac{1}{3} \tan^{- 1} \frac{x}{3}]}_{0}^{\infty} = \frac{1}{5} [\frac{1}{2} \times \frac{π}{2} - \frac{1}{3} \times \frac{π}{2}] = \frac{1}{5} \times \frac{π}{12} = \frac{π}{60}$

Page No 20.118:

Question 14:

$\int_{1}^{e} \log x d x =$

(a) 1
(b) e − 1
(c) e + 1
(d) 0

Answer:

(a) 1

$\int_{1}^{e} \log x d x = \int_{1}^{e} \log x x^{0} d x = {[x \log x]}_{1}^{e} - \int_{1}^{e} \frac{1}{x} x d x = {[x \log x]}_{1}^{e} - {[x]}_{1}^{e} = (e - 0) - (e - 1) = e - e + 1 = 1$

Page No 20.118:

Question 15:

$\int_{1}^{\sqrt{3}} \frac{1}{1 + x^{2}} d x$ is equal to

(a) $\frac{π}{12}$

(b) $\frac{π}{6}$

(c) $\frac{π}{4}$

(d) $\frac{π}{3}$

Answer:

(a) $\frac{π}{12}$

$\int_{1}^{\sqrt{3}} \frac{1}{1 + x^{2}} d x = {[\tan^{- 1} x]}_{1}^{\sqrt{3}} = \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$
30. →

Page No 20.118:

Question 16:

$\int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} d x =$

(a) $\frac{π}{12} + \log (2 \sqrt{2})$

(b) $\frac{π}{2} + \log (2 \sqrt{2})$

(c) $\frac{π}{6} + \log (2 \sqrt{2})$

(d) $\frac{π}{3} + \log (2 \sqrt{2})$

Answer:

$(a) \frac{π}{12} + \log (2 \sqrt{2})$

$We have, I = \int_{0}^{3} \frac{3 x + 1}{x^{2} + 9} d x I = \int_{0}^{3} \frac{3 x}{x^{2} + 9} d x + \int_{0}^{3} \frac{1}{x^{2} + 9} d x I_{1} = \int_{0}^{3} \frac{3 x}{x^{2} + 9} d x and I_{2} = \int_{0}^{3} \frac{1}{x^{2} + 9} d x Putting x^{2} + 9 = t in I_{1} \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} When x \to 0; t \to 9 and x \to 3; t \to 18 ∴ I = \int_{9}^{18} \frac{3 d t}{2 t} + \int_{0}^{3} \frac{1}{x^{2} + 9} d x = \frac{3}{2} \int_{9}^{18} \frac{d t}{t} + \int_{0}^{3} \frac{1}{x^{2} + 3^{2}} d x = \frac{3}{2} {[\log (t)]}_{9}^{18} + \frac{1}{3} {[\tan^{- 1} (\frac{x}{3})]}_{0}^{3} = \frac{3}{2} [\log 18 - \log 9] + \frac{1}{3} (\frac{π}{4} - 0) = \frac{3}{2} [\log \frac{18}{9}] + \frac{π}{12} = \frac{3}{2} [\log 2] + \frac{π}{12} = \log (\sqrt{8}) + \frac{π}{12} = \log (2 \sqrt{2}) + \frac{π}{12} = \frac{π}{12} + \log (2 \sqrt{2})$

Page No 20.118:

Question 17:

The value of the integral $\int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x$ is

(a) $\frac{π}{2}$

(b) $\frac{π}{4}$

(c) $\frac{π}{6}$

(d) $\frac{π}{3}$

Answer:

$(b) \frac{π}{4}$

$We have, I = \int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x Putting x = \tan θ \Rightarrow d x = \sec^{2} θ d θ When x \to 0; θ \to 0 and x \to \infty; θ \to \frac{π}{2} Now, integral becomes$

$I = \int_{0}^{\frac{π}{2}} \frac{\tan θ}{(1 + \tan θ) \sec^{2} θ} \sec^{2} θ d θ = \int_{0}^{\frac{π}{2}} \frac{\tan θ}{1 + \tan θ} d θ = \int_{0}^{\frac{π}{2}} \frac{\frac{\sin θ}{co s θ}}{1 + \frac{\sin θ}{\cos θ}} d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{\sin θ + \cos θ} d θ . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - θ)}{\sin (\frac{π}{2} - θ) + \cos (\frac{π}{2} - θ)} d θ [∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x] \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\cos θ + \sin θ} d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos θ}{\sin θ + \cos θ} d θ . . . . . (2)$

$Adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{2}} \frac{\sin θ + \cos θ}{\sin θ + \cos θ} d θ \Rightarrow 2 I = \int_{0}^{\frac{π}{2}} d θ \Rightarrow 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4} ∴ \int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x = \frac{π}{4}$

Page No 20.118:

Question 18:

$\int_{- π / 2}^{π / 2} \sin |x| d x$ is equal to

(a) 1
(b) 2
(c) − 1
(d) − 2

Answer:

(b) 2

$\int_{- \frac{π}{2}}^{\frac{π}{2}} \sin |x| d x = - \int_{- \frac{π}{2}}^{0} \sin x d x + \int_{0}^{\frac{π}{2}} \sin x d x = - {[- \cos x]}_{- \frac{π}{2}}^{0} + {[- \cos x]}_{0}^{\frac{π}{2}} = 1 - 0 - 0 + 1 = 2$

Page No 20.118:

Question 19:

$\int_{0}^{π / 2} \frac{1}{1 + \tan x} d x$ is equal to

(a) $\frac{π}{4}$

(b) $\frac{π}{3}$

(c) $\frac{π}{2}$

(d) π

Answer:

(a) $\frac{π}{4}$

$Let, I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t x} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{\frac{π}{2}} [\frac{1}{1 + \tan x} + \frac{1}{1 + c o t x}] d x = \int_{0}^{\frac{π}{2}} [\frac{(1 + c o t x) + (1 + \tan x)}{(1 + \tan x) (1 + c o t x)}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan x + \cot x}{1 + \tan x + c o t x + \tan x \cot x}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan x + \cot x}{2 + \tan x + \cot x}] d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence, I = \frac{π}{4}$

Page No 20.118:

Question 20:

The value of $\int_{0}^{π / 2} \cos x e^{\sin x} d x$ is
(a) 1
(b) e − 1
(c) 0
(d) − 1

Answer:

(b) e − 1

$Let, I = \int_{0}^{\frac{π}{2}} \cos x e^{\sin x} d x Let \sin x = t, then \cos x d x = d t When x = 0, t = 0 and x = \frac{π}{2}, t = 1 Therefore the integral becomes I = \int_{0}^{1} e^{t} dt = {[e^{t}]}_{0}^{1} = e - 1$

Page No 20.118:

Question 21:

If $\int_{0}^{a} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8},$ then a equals

(a) $\frac{π}{2}$

(b) $\frac{1}{2}$

(c) $\frac{π}{4}$

(d) 1

Answer:

(b) $\frac{1}{2}$

$\int_{0}^{α} \frac{1}{1 + 4 x^{2}} d x = \frac{π}{8} \Rightarrow \int_{0}^{α} \frac{1}{1 + {(2 x)}^{2}} d x = \frac{π}{8} \Rightarrow \frac{1}{2} {[\tan^{- 1} 2 x]}_{0}^{α} = \frac{π}{8} \Rightarrow \frac{1}{2} \tan^{- 1} 2 α = \frac{π}{8} \Rightarrow 2 α = \tan \frac{π}{4} \Rightarrow 2 α = 1 ∴ α = \frac{1}{2}$ 4430. →

Page No 20.118:

Question 22:

If $\int_{0}^{1} f (x) d x = 1, \int_{0}^{1} x f (x) d x = a, \int_{0}^{1} x^{2} f (x) d x = a^{2}, then \int_{0}^{1} {(a - x)}^{2} f (x) d x$ equals

(a) 4a²
(b) 0
(c) 2a²
(d) none of these

Answer:

(b) 0

$\int_{0}^{1} {(a - x)}^{2} f (x) d x = a^{2} \int_{0}^{1} f (x) d x + \int_{0}^{1} x^{2} f (x) d x - 2 a \int_{0}^{1} x f (x) d x = a^{2} \times 1 + a^{2} - 2 a a (As per given values) = 2 a^{2} - 2 a^{2} = 0$

Page No 20.119:

Question 23:

The value of $\int_{- π}^{π} \sin^{3} x \cos^{2} x d x$ is

(a) $\frac{π^{4}}{2}$

(b) $\frac{π^{4}}{4}$

(c) 0

(d) none of these

Answer:

(c) 0

$\int_{- π}^{π} \sin^{3} x \cos^{2} x d x = \int_{- π}^{π} \sin x (1 - \cos^{2} x) \cos^{2} x d x Let \cos x = t, then - \sin x d x = d t, When, x = - π, t = - 1, x = π, t = - 1 Therefore the integral becomes \int_{- 1}^{- 1} - (1 - t^{2}) t^{2} dt = 0$

Page No 20.119:

Question 24:

$\int_{π / 6}^{π / 3} \frac{1}{\sin 2 x} d x$ is equal to

(a) log_e 3

(b) $\log_{e} \sqrt{3}$

(c) $\frac{1}{2} \log (- 1)$

(d) log (−1)

Answer:

(b) $\log_{e} \sqrt{3}$

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{\sin 2 x} d x = \int_{\frac{π}{6}}^{\frac{π}{3}} \cos e c 2 x d x = \frac{1}{2} \int_{\frac{π}{6}}^{\frac{π}{3}} 2 \cos e c 2 x d x = \frac{- 1}{2} {[\log (\cos e c 2 x + c o t 2 x)]}_{\frac{π}{6}}^{\frac{π}{3}} = \frac{- 1}{2} [- 2 \log \sqrt{3}] = \log \sqrt{3}$

Page No 20.119:

Question 25:

$\int_{- 1}^{1} |1 - x| d x$ is equal to

(a) −2
(b) 2
(c) 0
(d) 4

Answer:

(b) 2

$\int_{- 1}^{1} |1 - x| d x = \int_{- 1}^{0} (1 - x) d x + \int_{0}^{1} (1 - x) d x = {[x - \frac{x^{2}}{2}]}_{- 1}^{0} + {[x - \frac{x^{2}}{2}]}_{0}^{1} = 0 + 1 + \frac{1}{2} + 1 - \frac{1}{2} - 0 = 2$

Page No 20.119:

Question 26:

The derivative of $f (x) = \int_{x^{2}}^{x^{3}} \frac{1}{\log_{e} t} d t, (x > 0),$ is

(a) $\frac{1}{3 \ln x}$

(b) $\frac{1}{3 \ln x} - \frac{1}{2 \ln x}$

(c) (ln x)⁻¹ x (x − 1)

(d) $\frac{3 x^{2}}{\ln x}$

Answer:

(c) (ln x)⁻¹ x (x − 1)

Using Newton Leibnitz formula

$\begin{matrix} f^{'} (x) = \frac{1}{\log_{e} x^{3}} (3 x^{2}) - \frac{1}{\log_{e} x^{2}} (2 x) \\ = \frac{3 x^{2}}{3 \ln x} - \frac{2 x}{2 \ln x} \\ = \frac{x^{2}}{\ln x} - \frac{x}{\ln x} \\ = \frac{1}{\ln x} x (x - 1) \\ = {(\ln x)}^{- 1} x (x - 1) \end{matrix}$

Page No 20.119:

Question 27:

If $I_{10} = \int_{0}^{π / 2} x^{10} \sin x d x,$ then the value of I₁₀ + 90I₈ is

(a) $9 {(\frac{π}{2})}^{9}$

(b) $10 {(\frac{π}{2})}^{9}$

(c) ${(\frac{π}{2})}^{9}$

(d) $9 {(\frac{π}{2})}^{8}$

Answer:

$(b) 10 {(\frac{π}{2})}^{9} We have, I_{10} = \int_{0}^{π / 2} x^{10} \sin x d x = {[x^{10} (- \cos x)]}_{0}^{\frac{π}{2}} - \int_{0}^{π / 2} [10 x^{9} \int \sin x d x] d x = {[- x^{10} \cos x]}_{0}^{\frac{π}{2}} - 10 \int_{0}^{π / 2} x^{9} (- \cos x) d x = - {[x^{10} \cos x]}_{0}^{\frac{π}{2}} + 10 \int_{0}^{π / 2} x^{9} \cos x d x = - {[x^{10} \cos x]}_{0}^{\frac{π}{2}} + 10 {[x^{9} \sin x]}_{0}^{\frac{π}{2}} - 10 \int_{0}^{π / 2} 9 x^{8} \sin x d x = - [{(\frac{π}{2})}^{10} \times 0 - 0^{10} \cos 0] + 10 [{(\frac{π}{2})}^{9} \times 1 - 0^{9} \times 0] - 90 \int_{0}^{π / 2} x^{8} \sin x d x = 10 [{(\frac{π}{2})}^{9} \times 1] - 90 I_{8} = 10 {(\frac{π}{2})}^{9} - 90 I_{8} ∴ I_{10} + 90 I_{8} = 10 {(\frac{π}{2})}^{9}$

Page No 20.119:

Question 28:

$\int_{0}^{1} \frac{x}{{(1 - x)}^{54}} d x =$

(a) $\frac{15}{16}$

(b) $\frac{3}{16}$

(c) $- \frac{3}{16}$

(d) $- \frac{16}{3}$

Answer:

Disclaimer: The question given is not correct because the function provided does not converge in the given domain.

Page No 20.119:

Question 29:

$\lim_{n □ \infty} \{\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + . . . + \frac{1}{2 n + n}\}$ is equal to

(a) $\ln (\frac{1}{3})$

(b) $\ln (\frac{2}{3})$

(c) $\ln (\frac{3}{2})$

(d) $\ln (\frac{4}{3})$

Answer:

(c) ln(3/2)

$\lim_{n \to \infty} \{\frac{1}{2 n + 1} + \frac{1}{2 n + 2} + . . . . . . . . . . + \frac{1}{2 n + n}\} = \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{2 n + r} = \underset{n \to \infty}{l i m} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{2 + \frac{r}{n}} l e t \frac{r}{n} = x = \int_{0}^{\infty} \frac{1}{2 + x} d x = {[\log (2 + x)]}_{0}^{\infty} = \log 3 - \log 2 = l o g \frac{3}{2} = \ln (\frac{3}{2})$

Page No 20.119:

Question 30:

The value of the integral $\int_{- 2}^{2} |1 - x^{2}| d x$ is
(a) 4
(b) 2
(c) −2
(d) 0

Answer:

(a) 4

$We have, I = \int_{- 2}^{2} |1 - x^{^{2}}| d x |1 - x^{^{2}}| = \{\begin{cases} \begin{array}{l} - (1 - x^{^{2}}), - 2 < x < - 1 \\ (1 - x^{^{2}}), - 1 < x < 1 \end{array} \\ - (1 - x^{^{2}}), 1 < x < 2 \end{cases} ∴ I = \int_{- 2}^{- 1} |1 - x^{^{2}}| d x + \int_{- 1}^{1} |1 - x^{^{2}}| d x + \int_{1}^{2} |1 - x^{^{2}}| d x = \int_{- 2}^{- 1} - (1 - x^{^{2}}) d x + \int_{- 1}^{1} (1 - x^{^{2}}) d x + \int_{1}^{2} - (1 - x^{^{2}}) d x = - \int_{- 2}^{- 1} (1 - x^{^{2}}) d x + \int_{- 1}^{1} (1 - x^{^{2}}) d x - \int_{1}^{2} (1 - x^{^{2}}) d x = - {[x - \frac{x^{3}}{3}]}_{- 2}^{- 1} + {[x - \frac{x^{3}}{3}]}_{- 1}^{1} - {[x - \frac{x^{3}}{3}]}_{1}^{2} = - [- 1 + \frac{1}{3} + 2 - \frac{8}{3}] + [1 - \frac{1}{3} + 1 - \frac{1}{3}] - [2 - \frac{8}{3} - 1 + \frac{1}{3}] = - [1 - \frac{7}{3}] + [2 - \frac{2}{3}] - [1 - \frac{7}{3}] = - 1 + \frac{7}{3} + 2 - \frac{2}{3} - 1 + \frac{7}{3} = 4$

Page No 20.119:

Question 31:

$\int_{0}^{π / 2} \frac{1}{1 + \cot^{3} x} d x$ is equal to

(a) 0
(b) 1
(c) π/2
(d) π/4

Answer:

(d) π/4

$We have, I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \cot^{3} x} d x . . . . . (1) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \cot^{3} (\frac{π}{2} - x)} d x ∴ I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} x} d x . . . . . (2) Adding (1) and (2) we get 2 I = \int_{0}^{\frac{π}{2}} [\frac{1}{1 + c o t^{3} x} + \frac{1}{1 + \tan^{3} x}] d x$

$= \int_{0}^{\frac{π}{2}} [\frac{1 + \tan^{3} x + 1 + co t^{3} x}{(1 + c o t^{3} x) (1 + \tan^{3} x)}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan^{3} x + co t^{3} x}{1 + \tan^{3} x + co t^{3} x + c o t^{3} x \tan^{3} x}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan^{3} x + co t^{3} x}{1 + \tan^{3} x + co t^{3} x + 1}] d x = \int_{0}^{\frac{π}{2}} [\frac{2 + \tan^{3} x + co t^{3} x}{2 + \tan^{3} x + co t^{3} x}] d x = \int_{0}^{\frac{π}{2}} [1] d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 20.119:

Question 32:

$\int_{0}^{π / 2} \frac{\sin x}{\sin x + \cos x} d x$ equals to

(a) π
(b) π/2
(c) π/3
(d) π/4

Answer:

(d) π/4

$We have, I = \int_{0}^{\frac{π}{2}} \frac{\sin x}{\sin x + \cos x} d x . . . . . (1) \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\cos x + \sin x} d x ∴ I = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\sin x + \cos x} d x . . . . . (2) Adding (1) and (2), we get 2 I = \int_{0}^{\frac{π}{2}} [\frac{\sin x}{\sin x + \cos x} + \frac{\cos x}{\cos x + \sin x}] d x = \int_{0}^{\frac{π}{2}} [\frac{\sin x + \cos x}{\sin x + \cos x}] d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence I = \frac{π}{4}$

Page No 20.120:

Question 33:

$\int_{0}^{1} \frac{d}{d x} \{\sin^{- 1} (\frac{2 x}{1 + x^{2}})\} d x$ is equal to

(a) 0
(b) π
(c) π/2
(d) π/4

Answer:

(c) π/2

$We have, I = \int_{0}^{1} \frac{d}{d x} \{\sin^{- 1} (\frac{2 x}{1 + x^{2}})\} d x We know since \int f' (x) = f (x) f (x) = s i n^{- 1} (\frac{2 x}{1 + x^{2}}) a n d f' (x) = \frac{d}{d x} \{s i n^{- 1} (\frac{2 x}{1 + x^{2}})\} Therefore, I = {[\sin^{- 1} (\frac{2 x}{1 + x^{2}})]}_{0}^{1} = \sin^{- 1} (1) - \sin^{- 1} (0) = \frac{π}{2}$

Page No 20.120:

Question 34:

$\int_{0}^{π / 2} x \sin x d x$ is equal to

(a) π/4
(b) π/2
(c) π
(d) 1

Answer:

(d) 1

$We have, I = \int_{0}^{\frac{π}{2}} x \sin x d x = {[- x \cos x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 (- \cos x) d x = {[- x \cos x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} \cos x d x = - {[x \cos x]}_{0}^{\frac{π}{2}} + {[\sin x]}_{0}^{\frac{π}{2}} = - [0 - 0] + [1 - 0] = 1$

Page No 20.120:

Question 35:

$\int_{0}^{π / 2} \sin 2 x \log \tan x d x$ is equal to

(a) π
(b) π/2
(c) 0
(d) 2π

Answer:

(c) 0

$I = \int_{0}^{\frac{π}{2}} \sin 2 x \log \tan x d x . . . . . (1) I = \int_{0}^{\frac{π}{2}} \sin (π - 2 x) \log \tan (\frac{π}{2} - x) d x I = \int_{0}^{\frac{π}{2}} \sin 2 x \log \cot x d x . . . . . (2) Adding (1) and (2), we get, 2 I = \int_{0}^{\frac{π}{2}} \sin 2 x (\log \tan x + \log \cot x) d x 2 I = \int_{0}^{\frac{π}{2}} \sin 2 x (\log \tan x \cot x) d x 2 I = \int_{0}^{\frac{π}{2}} \sin 2 x (\log 1) d x I = 0$

Page No 20.120:

Question 36:

The value of $\int_{0}^{π} \frac{1}{5 + 3 \cos x} d x$ is

(a) π/4
(b) π/8
(c) π/2
(d) 0

Answer:

(a) π/4

$\int_{0}^{π} \frac{1}{5 + 3 \cos x} d x = \int_{0}^{π} \frac{1}{5 + 3 \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{5 + 5 \tan^{2} \frac{x}{2} + 3 - 3 \tan^{2} \frac{x}{2}} d x = \int_{0}^{π} \frac{s e c^{2} \frac{x}{2}}{8 + 2 \tan^{2} \frac{x}{2}} d x Let \tan \frac{x}{2} = t, then s e c^{2} \frac{x}{2} d x = 2 d t When x = 0, t = 0, x = π, t = \infty Therefore the integral becomes \frac{1}{2} \int_{0}^{\infty} \frac{dt}{4 + t^{2}} = \frac{1}{2} {[\tan^{- 1} \frac{t}{2}]}_{0}^{\infty} = \frac{1}{2} (\frac{π}{2} - 0) = \frac{π}{4}$

Page No 20.120:

Question 37:

$\int_{0}^{\infty} \log (x + \frac{1}{x}) \frac{1}{1 + x^{2}} d x =$

(a) π ln 2
(b) −π ln 2
(c) 0
(d) $- \frac{π}{2} \ln 2$

Answer:

(a) π ln 2

$\int_{0}^{\infty} \log (x + \frac{1}{x}) \frac{1}{1 + x^{2}} d x$
Substitute x = tan θ
⇒ dx = sec² θ dθ.
when,
x = 0 ⇒ θ = 0
$x = \infty \Rightarrow θ = \frac{π}{2} \int_{0}^{\frac{π}{2}} (\tan θ + \frac{1}{\tan θ}) \frac{1}{1 + \tan^{2} θ} \times \sec^{2} θ d θ \int_{0}^{\frac{π}{2}} \log (\frac{\tan^{2} θ + 1}{\tan θ}) \frac{1}{1 + \tan^{2} θ} \times \sec^{2} θ d θ \Rightarrow \int_{0}^{\frac{π}{2}} \log (\frac{\sec^{2} θ}{\tan θ}) \frac{1}{\sec^{2} θ} \times \sec^{2} θ d θ [∵ 1 + \tan^{2} θ = \sec^{2} θ] \Rightarrow \int_{0}^{\frac{π}{2}} \log (\frac{\sec^{2} θ}{\tan θ}) d θ \Rightarrow \int_{0}^{\frac{π}{2}} \log (\frac{1}{\sin θ . \cos θ}) d θ \Rightarrow - \int_{0}^{\frac{π}{2}} \log (\sin θ . \cos θ) d θ \Rightarrow - \int_{0}^{\frac{π}{2}} [\log \sin θ + \log \cos θ] d θ \Rightarrow - \int_{0}^{\frac{π}{2}} \log \sin θ d θ - \int_{0}^{\frac{π}{2}} \log \cos θ d θ$
Let us consider,
$\int_{0}^{\frac{π}{2}} \log \sin θ d θ = I . . . . . (i) \Rightarrow I = \int_{0}^{\frac{π}{2}} \log (\sin (\frac{π}{2} - θ)) d θ = \int_{0}^{\frac{π}{2}} \log \cos θ d θ . . . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} \log \sin θ d θ + \int_{0}^{\frac{π}{2}} \log \cos θ d θ = \int_{0}^{\frac{π}{2}} \log (\sin θ . \cos θ) d θ = \int_{0}^{\frac{π}{2}} \log (\sin 2 θ) d θ - \int_{0}^{\frac{π}{2}} \log 2 d θ Let us consider 2 θ = t 2 d θ = d t 2 I = \frac{1}{2} \int_{0}^{π} \log (\sin t) d t - \frac{π}{2} \log 2 2 I = \frac{2}{2} \int_{0}^{\frac{π}{2}} \log (\sin t) d t - \frac{π}{2} \log 2 [∵ \sin θ is positive in both 1^{st} and 2^{nd} quadrants] 2 I = I - \frac{π}{2} \log 2 2 I - I = - \frac{π}{2} \log 2 I = - \frac{π}{2} \log 2, where I = \int_{0}^{\frac{π}{2}} \log \sin θ d θ Now, - \int_{0}^{\frac{π}{2}} \log (\sin θ) d θ - \int_{0}^{\frac{π}{2}} \log \cos θ d θ - 2 \int_{0}^{\frac{π}{2}} \log \sin θ d θ = - 2 \times I = - 2 \times - \frac{π}{2} \log 2 [∵ where I = - \frac{π}{2} \log 2] = π \log 2$

Page No 20.120:

Question 38:

$\int_{0}^{2 a} f (x) d x$ is equal to

(a) $2 \int_{0}^{a} f (x) d x$

(b) 0

(c) $\int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x$

(d) $\int_{0}^{a} f (x) d x + \int_{0}^{2 a} f (2 a - x) d x$

Answer:

$(c) \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x$

$A c c o r d i n g t o t h e a d d i t i v i t y p r o p e r t y o f i n t e g r a l s, \int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x, w h e r e a < c < b u \sin g t h i s p r o p e r t y, \int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{2 a} f (x) d x . . . . . . (1) N o w, c o n s i d e r t h e i n t e g r a l, \int_{0}^{2 a} f (x) d x L e t x = 2 a - t . T h e n d x = d (2 a - t) \Rightarrow d x = - d t A l s o, x = a \Rightarrow t = a a n d x = 2 a \Rightarrow t = 0 T h e r e f o r e, \int_{a}^{2 a} f (x) d x = - \int_{a}^{0} f (2 a - t) d t \Rightarrow \int_{a}^{2 a} f (x) d x = \int_{0}^{a} f (2 a - t) d t \Rightarrow \int_{a}^{2 a} f (x) d x = \int_{0}^{a} f (2 a - x) d x S u b s t i t u t i n g t h i s i n e q u a t i o n (1) w e g e t, \int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x$ →

Page No 20.120:

Question 39:

If f (a + b − x) = f (x), then $\int_{a}^{b}$ x f (x) dx is equal to

(a) $\frac{a + b}{2} \int_{a}^{b} f (b - x) d x$

(b) $\frac{a + b}{2} \int_{a}^{b} f (b + x) d x$

(c) $\frac{b - a}{2} \int_{a}^{b} f (x) d x$

(d) $\frac{a + b}{2} \int_{a}^{b} f (x) d x$

Answer:

(d) $\frac{a + b}{2} \int_{a}^{b} f (x) d x$

$Let, I = \int_{a}^{b} x f (x) d x . . . (i) = \int_{a}^{b} (a + b - x) f (a + b - x) d x = \int_{a}^{b} (a + b - x) f (x) d x . . . (ii) Adding (i) and (ii) 2 I = \int_{a}^{b} (x + a + b - x) f (x) d x = (a + b) \int_{a}^{b} f (x) d x Hence I = \frac{a + b}{2} \int_{a}^{b} f (x) d x$

Page No 20.120:

Question 40:

The value of $\int_{0}^{1} \tan^{- 1} (\frac{2 x - 1}{1 + x - x^{2}}) d x,$ is

(a) 1
(b) 0
(c) −1
(d) π/4

Answer:

(b) 0

$Let, I = \int_{0}^{1} \tan^{- 1} \frac{2 x - 1}{1 + x - x^{2}} d x . . . (i) = \int_{0}^{1} \tan^{- 1} \frac{2 (1 - x) - 1}{1 + (1 - x) - {(1 - x)}^{2}} d x = \int_{0}^{1} \tan^{- 1} \frac{1 - 2 x}{2 - x - 1 - x^{2} + 2 x} d x = \int_{0}^{1} \tan^{- 1} \frac{1 - 2 x}{1 + x - x^{2}} d x = - \int_{0}^{1} \tan^{- 1} \frac{2 x - 1}{1 + x - x^{2}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{1} \tan^{- 1} \frac{2 x - 1}{1 + x - x^{2}} d x - \int_{0}^{1} \tan^{- 1} \frac{2 x - 1}{1 + x - x^{2}} d x = 0 Hence, I = 0$

Page No 20.120:

Question 41:

The value of $\int_{0}^{π / 2} \log (\frac{4 + 3 \sin x}{4 + 3 \cos x}) d x$ is

(a) 2

(b) $\frac{3}{4}$

(c) 0

(d) −2

Answer:

(c) 0

$Let I = \int_{0}^{\frac{π}{2}} \log (\frac{4 + 3 \sin x}{4 + 3 \cos x}) d x . . . (i) = \int_{0}^{\frac{π}{2}} \log [\frac{4 + 3 \sin (\frac{π}{2} - x)}{4 + 3 \cos (\frac{π}{2} - x)}] d x = \int_{0}^{\frac{π}{2}} \log (\frac{4 + 3 \cos x}{4 + 3 \sin x}) d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} [\log (\frac{4 + 3 \sin x}{4 + 3 \cos x}) + l og (\frac{4 + 3 \cos x}{4 + 3 \sin x})] d x = \int_{0}^{\frac{π}{2}} \log (\frac{4 + 3 \sin x}{4 + 3 \cos x} \times \frac{4 + 3 \cos x}{4 + 3 \sin x}) d x = \int_{0}^{\frac{π}{2}} \log 1 d x = 0 Hence I = 0$

Page No 20.120:

Question 42:

The value of $\int_{- π / 2}^{π / 2} (x^{3} + x \cos x + \tan^{5} x + 1) d x,$ is

(a) 0
(b) 2
(c) π
(d) 1

Answer:

(c) $π$

$\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{3} + x \cos x + \tan^{5} x + 1) d x = {[\frac{x^{4}}{4}]}_{- \frac{π}{2}}^{\frac{π}{2}} + {[x \sin x]}_{- \frac{π}{2}}^{\frac{π}{2}} - \int_{- \frac{π}{2}}^{\frac{π}{2}} \sin x d x + \int_{- \frac{π}{2}}^{\frac{π}{2}} \tan^{3} x (s e c^{2} x - 1) d x + {[x]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{π^{4}}{64} - \frac{π^{4}}{64} + \frac{π}{2} - \frac{π}{2} - {[- \cos x]}_{- \frac{π}{2}}^{\frac{π}{2}} + \int_{- \frac{π}{2}}^{\frac{π}{2}} \tan^{3} x s e c^{2} x d x - \int_{- \frac{π}{2}}^{\frac{π}{2}} \tan^{3} x d x + \frac{π}{2} + \frac{π}{2} = π + 0 + {[\frac{\tan^{4} x}{4}]}_{- \frac{π}{2}}^{\frac{π}{2}} - \int_{- \frac{π}{2}}^{\frac{π}{2}} tanx \sec^{2} x dx - \int_{- \frac{π}{2}}^{\frac{π}{2}} tanx dx = π - {[\frac{\tan^{2} x}{2}]}_{- \frac{π}{2}}^{\frac{π}{2}} - {[- \log (\cos x)]}_{- \frac{π}{2}}^{\frac{π}{2}} = π$

Page No 20.121:

Question 1:

$\int_{0}^{4} x \sqrt{4 - x} d x$

Answer:

$Let, I = \int_{0}^{4} x \sqrt{4 - x} d x = \int_{0}^{4} (4 - x) \sqrt{4 - 4 + x} d x = \int_{0}^{4} (4 - x) \sqrt{x} d x = \int_{0}^{4} 4 \sqrt{x} - x^{\frac{3}{2}} d x = {[8 \frac{x^{\frac{3}{2}}}{3}]}_{0}^{4} - {[\frac{2 x^{\frac{5}{2}}}{5}]}_{0}^{4} = \frac{64}{3} - \frac{64}{5} = \frac{128}{15}$ →

Page No 20.121:

Question 2:

$\int_{1}^{2} x \sqrt{3 x - 2} d x$

Answer:

$\int_{1}^{2} x \sqrt{3 x - 2} d x Let, 3 x - 2 = t, then 3 d x = d t when, x = 1; t = 1 and x = 2; t = 4 Therefore the integral becomes \int_{1}^{4} \frac{t + 2}{3} \sqrt{t} \frac{d t}{3} = \frac{1}{9} \int_{1}^{4} t^{\frac{3}{2}} + 2 \sqrt{t} d t = \frac{1}{9} {[\frac{2 t^{\frac{5}{2}}}{5} + \frac{4 t^{\frac{3}{2}}}{3}]}_{1}^{4} = \frac{1}{9} [\frac{64}{5} + \frac{32}{3} - \frac{2}{5} - \frac{4}{3}] = \frac{46}{135}$

Page No 20.121:

Question 3:

$\int_{1}^{5} \frac{x}{\sqrt{2 x - 1}} d x$

Answer:

$Let I = \int_{1}^{5} \frac{x}{\sqrt{2 x - 1}} d x Let, 2 x - 1 = t, then 2 d x = d t, When, x \to 1; t \to 1 and x \to 5; t \to 9 x = \frac{t + 1}{2} I = \frac{1}{2} \int_{1}^{9} \frac{t + 1}{\sqrt{t}} \times \frac{d t}{2} = \frac{1}{4} {[\frac{2 t^{\frac{3}{2}}}{3} + 2 \sqrt{t}]}_{1}^{9} = \frac{1}{4} [18 + 6 - \frac{2}{3} - 2] = \frac{16}{3}$

Page No 20.121:

Question 4:

$\int_{0}^{1} \cos^{- 1} x d x$

Answer:

$\int_{0}^{1} \cos^{- 1} x d x = \int_{0}^{1} (\cos^{- 1} x \times 1) d x = {[\cos^{- 1} x x]}_{0}^{1} - \int_{0}^{1} \frac{- x}{\sqrt{1 - x^{2}}} d x = {[x \cos^{- 1} x]}_{0}^{1} - \frac{2}{2} {[\sqrt{1 - x^{2}}]}_{0}^{1} = 0 + 1 = 1$

Page No 20.121:

Question 5:

$\int_{0}^{1} \tan^{- 1} x d x$

Answer:

$\int_{0}^{1} \tan^{- 1} x d x = \int_{0}^{1} \tan^{- 1} x \times 1 d x = {[\tan^{- 1} x x]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x^{2}} d x = {[x \tan^{- 1} x]}_{0}^{1} - \frac{1}{2} {[\log (1 + x^{2})]}_{0}^{1} = \frac{π}{4} - 0 - \frac{1}{2} \log 2 + 0 = \frac{π}{4} - \frac{1}{2} \log 2$

Page No 20.121:

Question 6:

$\int_{0}^{1} \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x$

Answer:

$\int_{0}^{1} \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) d x Let, x = \tan θ, d x = s e c^{2} θ d θ When, x \to 0; θ \to 0 and x \to 1; θ \to \frac{π}{4} Therefore, the integral becomes \int_{0}^{\frac{π}{4}} \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) s e c^{2} θ d θ = \int_{0}^{\frac{π}{4}} \cos^{- 1} (\cos 2 θ) s e c^{2} θ d θ = 2 \int_{0}^{\frac{π}{4}} θ s e c^{2} θ d θ = 2 {[θ t an θ]}_{0}^{\frac{π}{4}} - 2 \int_{0}^{\frac{π}{4}} \tan θ d θ = 2 {[θ \tan θ]}_{0}^{\frac{π}{4}} + 2 {[\log (\cos θ)]}_{0}^{\frac{π}{4}} = 2 (\frac{π}{4} - 0) + 2 [\log \frac{1}{\sqrt{2}} - 0] = \frac{π}{2} - \log 2$

Page No 20.121:

Question 7:

$\int_{0}^{1} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x$

Answer:

$\int_{0}^{1} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) d x Let, x = \tan θ, then d x = s e c^{2} θ d θ When, x \to 0; θ \to 0 And x \to 1; θ \to \frac{π}{4} Therefore the integral becomes \int_{0}^{\frac{π}{4}} \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) s e c^{2} θ d θ = \int_{0}^{\frac{π}{4}} \tan^{- 1} (\tan 2 θ) s e c^{2} θ d θ = 2 \int_{0}^{\frac{π}{4}} θ s e c^{2} θ d θ = 2 {[θ \tan θ]}_{0}^{\frac{π}{4}} - 2 \int_{0}^{\frac{π}{4}} \tan θ d θ = 2 {[θ \tan θ]}_{0}^{\frac{π}{4}} - 2 {[- \log (\cos θ)]}_{0}^{\frac{π}{4}} = 2 (\frac{π}{4} - 0) + 2 [\log \frac{1}{\sqrt{2}} - 0] = \frac{π}{2} - \log 2$

Page No 20.121:

Question 8:

$\int_{0}^{1 / \sqrt{3}} \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) d x$

Answer:

$\int_{0}^{\frac{1}{\sqrt{3}}} \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) d x Let x = \tan θ, then d x = s e c^{2} θ d θ When, x \to 0; θ \to 0 And x \to \frac{1}{\sqrt{3}}; θ \to \frac{π}{6} Therefore the integral becomes \int_{0}^{\frac{π}{6}} \tan^{- 1} (\frac{3 \tan θ - \tan^{3} θ}{1 - 3 \tan^{2} θ}) s e c^{2} θ d θ = \int_{0}^{\frac{π}{6}} \tan^{- 1} (\tan 3 θ) s e c^{2} θ d θ = 3 \int_{0}^{\frac{π}{6}} θ s e c^{2} θ d θ = 3 {[θ \tan θ]}_{0}^{\frac{π}{6}} - 3 \int_{0}^{\frac{π}{6}} \tan θ d θ = 3 {[θ \tan θ]}_{0}^{\frac{π}{6}} - 3 {[- \log (\cos θ)]}_{0}^{\frac{π}{6}} = 3 (\frac{π}{6} \times \frac{1}{\sqrt{3}} - 0) + 3 [\log \frac{\sqrt{3}}{2}] = \frac{π}{2 \sqrt{3}} + 3 \log \frac{\sqrt{3}}{2} = \frac{π}{2 \sqrt{3}} - \frac{3}{2} \log \frac{4}{3}$

Page No 20.121:

Question 9:

$\int_{0}^{1} \frac{1 - x}{1 + x} d x$

Answer:

$\int_{0}^{1} \frac{1 - x}{1 + x} d x = \int_{0}^{1} \frac{1 - x - 1 + 1}{1 + x} d x = \int_{0}^{1} \frac{2 - (x + 1)}{1 + x} d x = \int_{0}^{1} \frac{2}{1 + x} - \int_{0}^{1} \frac{1 + x}{1 + x} d x = \int_{0}^{1} \frac{2}{1 + x} - \int_{0}^{1} d x = 2 {[\log (1 + x)]}_{0}^{1} - {[x]}_{0}^{1} = 2 \log 2 - 1$

Page No 20.121:

Question 10:

$\int_{0}^{π / 3} \frac{\cos x}{3 + 4 \sin x} d x$

Answer:

$\int_{0}^{\frac{π}{3}} \frac{\cos x}{3 + 4 \sin x} d x Let, \sin x = t \Rightarrow \cos x d x = d t When, \sin x \to 0; t \to 0 And \sin x \to \frac{π}{3}; t \to \frac{\sqrt{3}}{2} = \int_{0}^{\frac{\sqrt{3}}{2}} \frac{d t}{3 + 4 t} = \frac{1}{4} \log {[3 + 4 t]}_{0}^{\frac{\sqrt{3}}{2}} = \frac{1}{4} \log [\log (3 + 2 \sqrt{3}) - \log (3 + 0)] = \frac{1}{4} \log [\log (2 \sqrt{3} + 3) - \log (3)] = \frac{1}{4} (\log \frac{2 \sqrt{3} + 3}{3})$

Page No 20.121:

Question 11:

$\int_{0}^{π / 2} \frac{\sin^{2} x}{{(1 + \cos x)}^{2}} d x$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{{(1 + \cos x)}^{2}} d x = \int_{0}^{\frac{π}{2}} \frac{1 - \cos^{2} x}{{(1 + \cos x)}^{2}} d x = \int_{0}^{\frac{π}{2}} \frac{(1 + \cos x) (1 - \cos x)}{{(1 + \cos x)}^{2}} d x = \int_{0}^{\frac{π}{2}} \frac{1 - \cos x}{1 + \cos x} d x = \int_{0}^{\frac{π}{2}} \frac{1 - \cos x - 1 + 1}{(1 + \cos x)} d x = \int_{0}^{\frac{π}{2}} \frac{2 - (1 + \cos x)}{(1 + \cos x)} d x = \int_{0}^{\frac{π}{2}} \frac{2}{1 + \cos x} d x - \int_{0}^{\frac{π}{2}} d x = \int_{0}^{\frac{π}{2}} \frac{2 (1 - \cos x)}{(1 + \cos x) (1 - \cos x)} d x - \int_{0}^{\frac{π}{2}} d x = 2 \int_{0}^{\frac{π}{2}} \frac{1 - \cos x}{\sin^{2} x} d x - {[x]}_{0}^{\frac{π}{2}} = 2 \int_{0}^{\frac{π}{2}} (\cos e c^{2} x - \cos e c x c o t x) d x - {[x]}_{0}^{\frac{π}{2}} = 2 {[- co t x + \cos e c x]}_{0}^{\frac{π}{2}} - {[x]}_{0}^{\frac{π}{2}} = 2 - \frac{π}{2}$

Page No 20.121:

Question 12:

$\int_{0}^{π / 2} \frac{\sin x}{\sqrt{1 + \cos x}} d x$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{\sin x}{\sqrt{1 + \cos x}} d x Let 1 + \cos x = t, then - \sin x d x = d t When, x \to 0, t \to 2 and x \to \frac{π}{2}, t \to 1 Therefore, the integral becomes \int_{2}^{1} \frac{- 1}{\sqrt{t}} d t = \int_{1}^{2} \frac{1}{\sqrt{t}} d t = 2 {[\sqrt{t}]}_{1}^{2} = 2 (\sqrt{2} - 1)$

Page No 20.121:

Question 13:

$\int_{0}^{π / 2} \frac{\cos x}{1 + \sin^{2} x} d x$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{\cos x}{1 + \sin^{2} x} d x Let \sin x = t, then \cos x d x = d t When x \to 0; t \to 0 And x \to \frac{π}{2}; t \to 1 Therefore the integral becomes \int_{0}^{1} \frac{d t}{1 + t^{2}} = {[\tan^{- 1} x]}_{0}^{1} = \frac{π}{4}$

Page No 20.121:

Question 14:

$\int_{0}^{π} \sin^{3} x (1 + 2 \cos x) {(1 + \cos x)}^{2} d x$

Answer:

$We have, I = \int_{0}^{π} \sin^{3} x (1 + 2 \cos x) {(1 + \cos x)}^{2} d x = \int_{0}^{π} \sin^{2} x (1 + 2 \cos x) {(1 + \cos x)}^{2} \sin x d x = \int_{0}^{π} (1 - \cos^{2} x) (1 + 2 \cos x) {(1 + \cos x)}^{2} \sin x d x Putting \cos x = t \Rightarrow - \sin x d x = d t When x \to 0; t \to 1 and x \to π; t \to - 1 ∴ I = - \int_{1}^{- 1} (1 - t^{2}) (1 + 2 t) {(1 + t)}^{2} d t = \int_{- 1}^{1} (1 - t^{2}) (1 + 2 t) {(1 + t)}^{2} d t = \int_{- 1}^{1} (1 + 2 t - t^{2} - 2 t^{3}) (1 + 2 t + t^{2}) d t = \int_{- 1}^{1} (1 + 2 t + t^{2} + 2 t + 4 t^{2} + 2 t^{3} - t^{2} - 2 t^{3} - t^{4} - 2 t^{3} - 4 t^{4} - 2 t^{5}) d t = \int_{- 1}^{1} (1 + 4 t + 4 t^{2} - 2 t^{3} - 5 t^{4} - 2 t^{5}) d t = {[t + 2 t^{2} + \frac{4 t^{3}}{3} - \frac{t^{4}}{2} - t^{5} - \frac{t^{6}}{3}]}_{- 1}^{1} = 1 + 2 + \frac{4}{3} - \frac{1}{2} - 1 - \frac{1}{3} - (- 1) - 2 {(- 1)}^{2} - \frac{4 {(- 1)}^{3}}{3} + \frac{{(- 1)}^{4}}{2} + {(- 1)}^{5} + \frac{{(- 1)}^{6}}{3} = 1 + 2 + \frac{4}{3} - \frac{1}{2} - 1 - \frac{1}{3} + 1 - 2 + \frac{4}{3} + \frac{1}{2} - 1 + \frac{1}{3} = \frac{8}{3}$

Page No 20.121:

Question 15:

$\int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x$

Answer:

I = $\int_{0}^{\infty} \frac{x}{(1 + x) (1 + x^{2})} d x$

using partial fraction,
$\frac{x}{(1 + x) (1 + x^{2})} \frac{A}{1 + x} + \frac{B x + C}{1 + x^{2}} x = A (1 + x^{2}) + (B x + C) (1 + x) x = A + A x^{2} + B x + B x^{2} + C + C x B + C = 1 A + C = 0 A + B = 0 s o, A = \frac{- 1}{2}, B = \frac{1}{2}, C = \frac{1}{2}$

putting the values of A,B and C we get

$\frac{\frac{- 1}{2}}{1 + x} + \frac{\frac{1}{2} x + \frac{1}{2}}{1 + x^{2}} = \frac{- 1}{2} [\frac{1}{1 + x}] + \frac{1}{2} [\frac{x + 1}{1 + x^{2}}] T h e r e f o r e, I = \int_{0}^{\infty} \frac{- 1}{2} [\frac{1}{1 + x}] + \frac{1}{2} [\frac{x + 1}{1 + x^{2}}] I = \frac{- 1}{2} {[\log |1 + x|]}_{0}^{\infty} + \frac{1}{2} \int_{0}^{\infty} [\frac{x}{1 + x^{2}} + \frac{1}{1 + x^{2}}] I = \frac{- 1}{2} {[l o g |1 + x|]}_{0}^{\infty} + \frac{1}{2 \times 2} \int_{0}^{\infty} [\frac{2 x}{1 + x^{2}}] + \frac{1}{2} \int_{0}^{\infty} \frac{1}{1 + x^{2}}$
$I = \frac{- 1}{2} {[\log |1 + x|]}_{0}^{\infty} + \frac{1}{4} {[\log |1 + x^{2}|]}_{0}^{\infty} + {[\frac{1}{2} t a n^{- 1} x]}_{0}^{\infty} I = \frac{- 1}{2} {[l o g |1 + x|]}_{0}^{\infty} + \frac{1}{2} \times \frac{1}{2} {[l o g |1 + x^{2}|]}_{0}^{\infty} + {[\frac{1}{2} t a n^{- 1} x]}_{0}^{\infty} I = \frac{1}{2} {[\log \frac{\sqrt{x^{2} + 1}}{x + 1}]}_{0}^{\infty} + {[\frac{1}{2} t a n^{- 1} x]}_{0}^{\infty} I = \frac{1}{2} {[l o g \frac{\sqrt{1 + \frac{1}{x^{2}}}}{1 + \frac{1}{x}}]}_{0}^{\infty} + {[\frac{1}{2} t a n^{- 1} x]}_{0}^{\infty} I = \frac{1}{2} [0] + \frac{1}{2} [t a n^{- 1} \infty - t a n^{- 1} 0]$
$I = \frac{π}{4}$

Page No 20.121:

Question 16:

$\int_{0}^{π / 4} \sin 2 x \sin 3 x d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{4}} \sin 2 x \sin 3 x d x . . . (i) \Rightarrow I = {[- \sin 2 x \frac{\cos 3 x}{3}]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} 2 \cos 2 x \frac{\cos 3 x}{3} d x \Rightarrow I = {[- \sin 2 x \frac{\cos 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{2}{3} {[\cos 2 x \frac{\sin 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{4}{9} \int_{0}^{\frac{π}{4}} \sin 2 x \sin 3 x d x \Rightarrow I = {[- \sin 2 x \frac{\cos 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{2}{3} {[\cos 2 x \frac{\sin 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{4}{9} I [From (i)] \Rightarrow \frac{5}{9} I = {[- \sin 2 x \frac{\cos 3 x}{3}]}_{0}^{\frac{π}{4}} + \frac{2}{3} {[\cos 2 x \frac{\sin 3 x}{3}]}_{0}^{\frac{π}{4}} \Rightarrow \frac{5}{9} I = \frac{1}{3 \sqrt{2}} + 0 \Rightarrow \frac{5}{9} I = \frac{1}{3 \sqrt{2}} ∴ I = \frac{3}{5 \sqrt{2}}$

Page No 20.121:

Question 17:

$\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x$

Answer:

$\int_{0}^{1} \sqrt{\frac{1 - x}{1 + x}} d x = \int_{0}^{1} \sqrt{\frac{1 - x}{1 + x} \times \frac{1 - x}{1 - x}} d x = \int_{0}^{1} \frac{1 - x}{\sqrt{1 - x^{2}}} d x = \int_{0}^{1} \frac{1}{\sqrt{1 - x^{2}}} d x - \int_{0}^{1} \frac{x}{\sqrt{1 - x^{2}}} d x = {[\sin^{- 1} x]}_{0}^{1} + {[\sqrt{1 - x^{2}}]}_{0}^{1} = \frac{π}{2} - 1$

Page No 20.121:

Question 18:

$\int_{1}^{2} \frac{1}{x^{2}} e^{- 1 / x} d x$

Answer:

$\int_{1}^{2} \frac{1}{x^{2}} e^{\frac{- 1}{x}} d x Let \frac{- 1}{x} = t, then \frac{1}{x^{2}} d x = d t When, x \to 1; t \to - 1 And x \to 2; t \to \frac{- 1}{2} Therefore the integral becomes \int_{- 1}^{\frac{- 1}{2}} e^{t} d t = {[e^{t}]}_{- 1}^{\frac{- 1}{2}} = e^{\frac{- 1}{2}} - e^{- 1} = \frac{\sqrt{e} - 1}{e}$

Page No 20.121:

Question 19:

$\int_{0}^{π / 4} \cos^{4} x \sin^{3} x d x$

Answer:

$\int_{0}^{\frac{π}{4}} \cos^{4} x \sin^{3} x d x = \int_{0}^{\frac{π}{4}} \cos^{4} x \sin x (1 - \cos^{2} x) d x = \int_{0}^{\frac{π}{4}} \cos^{4} x \sin x d x - \int_{0}^{\frac{π}{4}} \cos^{6} x \sin x d x = - {[\frac{\cos^{5} x}{5}]}_{0}^{\frac{π}{4}} + {[\frac{\cos^{7} x}{7}]}_{0}^{\frac{π}{4}} = \frac{- 1}{20 \sqrt{2}} + \frac{1}{5} + \frac{1}{56 \sqrt{2}} - \frac{1}{7} = \frac{- \sqrt{2}}{40} + \frac{2}{35} + \frac{\sqrt{2}}{112} = \frac{2}{35} - \frac{9 \sqrt{2}}{560}$

Page No 20.121:

Question 20:

$\int_{π / 3}^{π / 2} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{5 / 2}} d x$

Answer:

$\int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{5}{2}}} d x = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sqrt{1 + \cos x}}{{(1 - \cos x)}^{\frac{5}{2}}} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} d x = \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{\sin x}{{(1 - \cos x)}^{3}} d x = - \frac{1}{2} {[{(1 - \cos x)}^{- 2}]}_{\frac{π}{3}}^{\frac{π}{2}} = - \frac{1}{2} [1 - 4] = \frac{3}{2}$

Page No 20.121:

Question 21:

$\int_{0}^{π / 2} x^{2} \cos 2 x d x$

Answer:

$\int_{0}^{\frac{π}{2}} x^{2} \cos 2 x d x = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x \frac{\sin 2 x}{2} d x = {[x^{2} \frac{s i n 2 x}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} x \sin 2 x d x = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} - {[- x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} + [- \int_{0}^{\frac{π}{2}} \frac{\cos 2 x}{2} d x] = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} + {[x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{\cos 2 x}{2} d x = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} + {[x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} - \frac{1}{2} {[\frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} = 0 - \frac{π}{4} - 0 = \frac{- π}{4}$

Page No 20.121:

Question 22:

$\int_{0}^{1} \log (1 + x) d x$

Answer:

$\int_{0}^{1} \log (1 + x) d x = \int_{0}^{1} \log (1 + x) \times 1 d x = {[\log (1 + x) x]}_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x} d x = {[\log (1 + x) x]}_{0}^{1} - \int_{0}^{1} (1 - \frac{1}{1 + x}) d x = {[x \log (1 + x)]}_{0}^{1} - {[x - \log (1 + x)]}_{0}^{1} = \log 2 - 1 + \log 2 = 2 \log 2 - 1 = \log 4 - \log e = \log \frac{4}{e}$

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Question 23:

Evaluate the following integrals:

$\int_{2}^{4} \frac{x^{2} + x}{\sqrt{2 x + 1}} d x$

Answer:

Let I = $\int_{2}^{4} \frac{x^{2} + x}{\sqrt{2 x + 1}} d x$

Put 2x + 1 = z^{2

$\Rightarrow 2 d x = 2 z d z \Rightarrow d x = z d z$}

When $x \to 2, z \to \sqrt{5}$

When $x \to 4, z \to 3$

$∴ I = \int_{\sqrt{5}}^{3} \frac{{(\frac{z^{2} - 1}{2})}^{2} + \frac{z^{2} - 1}{2}}{z} \times z d z \Rightarrow I = \int_{\sqrt{5}}^{3} \frac{z^{4} - 2 z^{2} + 1 + 2 z^{2} - 2}{4} d z \Rightarrow I = \frac{1}{4} \int_{\sqrt{5}}^{3} (z^{4} - 1) d z \Rightarrow I = \frac{1}{4} \times {(\frac{z^{5}}{5} - z)}_{\sqrt{5}}^{3}$
$\Rightarrow I = \frac{1}{4} [(\frac{243}{5} - 3) - (\frac{25 \sqrt{5}}{5} - \sqrt{5})] \Rightarrow I = \frac{1}{4} \times \frac{228}{5} - \frac{1}{4} \times 4 \sqrt{5} \Rightarrow I = \frac{57}{5} - \sqrt{5}$

Page No 20.121:

Question 24:

$\int_{0}^{1} x {(\tan^{- 1} x)}^{2} d x$

Answer:

$We have, I = \int_{0}^{1} x {(\tan^{- 1} x)}^{2} d x Putting \tan^{- 1} x = u \Rightarrow x = \tan u \Rightarrow d x = \sec^{2} u d u When x \to 0; u \to 0 and x \to 1; u \to \frac{π}{4} ∴ I = \int_{0}^{\frac{π}{4}} (\tan u) u^{2} \sec^{2} u d u = \int_{0}^{\frac{π}{4}} u^{2} \tan u \sec^{2} u d u = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} 2 u \frac{\tan^{2} u}{2} d u = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} u (\sec^{2} u - 1) d u = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} u \sec^{2} u d u + \int_{0}^{\frac{π}{4}} u d u = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - {[u \tan u]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} \tan u d u + {[\frac{u^{2}}{2}]}_{0}^{\frac{π}{4}} = {[u^{2} \frac{\tan^{2} u}{2}]}_{0}^{\frac{π}{4}} - {[u \tan u]}_{0}^{\frac{π}{4}} + {[\log |\sec u|]}_{0}^{\frac{π}{4}} + {[\frac{u^{2}}{2}]}_{0}^{\frac{π}{4}} = \frac{π^{2}}{16} \times \frac{1}{2} - \frac{π}{4} + \log \sqrt{2} + \frac{π^{2}}{32} = \frac{π^{2}}{16} - \frac{π}{4} + \log \sqrt{2} = \frac{π^{2}}{16} - \frac{π}{4} + \frac{1}{2} \log 2$

Page No 20.121:

Question 25:

$\int_{0}^{1} {(\cos^{- 1} x)}^{2} d x$

Answer:

$I = \int_{0}^{1} (\cos^{- 1} x)^{2} d x l e t c o s^{- 1} x = θ \Rightarrow x = \cos θ \Rightarrow d x = - \sin θ d θ w h e n x = 0, θ = \frac{π}{2} a n d w h e n x = 1, θ = 0 t h e r e f o r e, I = \int_{\frac{π}{2}}^{0} θ^{2} (- \sin θ) d θ I = - \int_{\frac{π}{2}}^{0} θ^{2} (s i n θ) d θ I = \int_{0}^{\frac{π}{2}} θ^{2} (s i n θ) d θ I = [θ^{2} (- c o s θ)]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 θ \int_{0}^{\frac{π}{2}} \sin θ d θ] I = [θ^{2} (- \cos θ)]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} [2 θ (- \cos θ) d θ]$
$= [- θ^{2} \cos θ]_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} 2 θ (\cos θ) d θ = [- θ^{2} c o s θ]_{0}^{\frac{π}{2}} + 2 [θ \sin θ - \int_{0}^{\frac{π}{2}} \sin θ d θ] = [- θ^{2} c o s θ]_{0}^{\frac{π}{2}} + 2 [θ s i n θ + \cos θ]_{0}^{\frac{π}{2}}$

$I = 2 [(\frac{π}{2} + 0) - 1] I = π - 2$

Page No 20.121:

Question 26:

$\int_{1}^{2} \frac{x + 3}{x (x + 2)} d x$

Answer:

$\int_{1}^{2} \frac{x + 3}{x (x + 2)} d x = \int_{1}^{2} \frac{x + 2 + 1}{x (x + 2)} d x = \int_{1}^{2} \frac{1}{x} d x + \int_{1}^{2} \frac{1}{x (x + 2)} d x = \int_{1}^{2} \frac{1}{x} d x + \frac{1}{2} \int_{1}^{2} \frac{(x + 2) - x}{x (x + 2)} d x = \int_{1}^{2} \frac{1}{x} d x + \frac{1}{2} \int_{1}^{2} \frac{1}{x} d x - \frac{1}{2} \int_{1}^{2} \frac{1}{x + 2} d x = \frac{3}{2} \int_{1}^{2} \frac{1}{x} d x - \frac{1}{2} \int_{1}^{2} \frac{1}{x + 2} d x = \frac{3}{2} {[\log x]}_{1}^{2} - \frac{1}{2} {[\log (x + 2)]}_{1}^{2} = \frac{3}{2} \log 2 - \frac{1}{2} \log 4 + \frac{1}{2} \log 3 = \frac{3}{2} \log 2 - \log 2 + \frac{1}{2} \log 3 = \frac{1}{2} \log 2 + \frac{1}{2} \log 3 = \frac{1}{2} \log 6$

Page No 20.122:

Question 27:

$\int_{0}^{π / 4} e^{x} \sin x d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{4}} e^{x} \sin x d x . . . (i) = {[- e^{x} \cos x]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} e^{x} \cos x d x = {[- e^{x} \cos x]}_{0}^{\frac{π}{4}} + {[e^{x} \sin x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} e^{x} \sin x d x \Rightarrow I = {[- e^{x} \cos x]}_{0}^{\frac{π}{4}} + {[e^{x} \sin x]}_{0}^{\frac{π}{4}} - I [Using (i)] \Rightarrow 2 I = {[- e^{x} \cos x]}_{0}^{\frac{π}{4}} + {[e^{x} \sin x]}_{0}^{\frac{π}{4}} = - \frac{1}{\sqrt{2}} e^{\frac{π}{4}} + 1 + \frac{1}{\sqrt{2}} e^{\frac{π}{4}} - 0 = 1 Hence I = \frac{1}{2}$

Page No 20.122:

Question 28:

$\int_{0}^{π / 4} \tan^{4} x d x$

Answer:

$\int_{0}^{\frac{π}{4}} \tan^{4} x d x = \int_{0}^{\frac{π}{4}} \tan^{2} x (s e c^{2} x - 1) d x = \int_{0}^{\frac{π}{4}} \tan^{2} x s e c^{2} x d x - \int_{0}^{\frac{π}{4}} \tan^{2} x d x = {[\frac{\tan^{3} x}{3}]}_{0}^{\frac{π}{4}} - {[\tan x - x]}_{0}^{\frac{π}{4}} = \frac{1}{3} - 1 + \frac{π}{4} = \frac{π}{4} - \frac{2}{3}$

Page No 20.122:

Question 29:

$\int_{0}^{1} |2 x - 1| d x$

Answer:

$We have, |2 x - 1| = \{\begin{cases} - (2 x - 1), 0 \leq x \leq \frac{1}{2} \\ 2 x - 1, \frac{1}{2} \leq x \leq 1 \end{cases} ∴ \int_{0}^{1} |2 x - 1| d x = \int_{0}^{\frac{1}{2}} - (2 x - 1) d x + \int_{\frac{1}{2}}^{1} (2 x - 1) d x = {[- x^{2} + x]}_{0}^{\frac{1}{2}} + {[x^{2} - x]}_{\frac{1}{2}}^{1} = \frac{- 1}{4} + \frac{1}{2} + 1 - 1 - \frac{1}{4} + \frac{1}{2} = \frac{1}{2}$

Page No 20.122:

Question 30:

$\int_{1}^{3} |x^{2} - 2 x| d x$

Answer:

$We have, |x^{2} - 2 x| = \{\begin{cases} - (x^{2} - 2 x), 1 \leq x \leq 2 \\ x^{2} - 2 x, 2 \leq x \leq 3 \end{cases} ∴ \int_{1}^{3} |x^{2} - 2 x| d x = \int_{1}^{2} - (x^{2} - 2 x) d x + \int_{2}^{3} (x^{2} - 2 x) d x = {[- \frac{x^{3}}{3} + x^{2}]}_{1}^{2} + {[\frac{x^{3}}{3} - x^{2}]}_{2}^{3} = \frac{- 8}{3} + 4 + \frac{1}{3} - 1 + 9 - 9 - \frac{8}{3} + 4 = 2$

Page No 20.122:

Question 31:

$\int_{0}^{π / 2} |\sin x - \cos x| d x$

Answer:

$\int_{0}^{\frac{π}{2}} |\sin x - \cos x| d x = \sqrt{2} \int_{0}^{\frac{π}{2}} |\sin x \frac{1}{\sqrt{2}} - \cos x \frac{1}{\sqrt{2}}| d x = \sqrt{2} \int_{0}^{\frac{π}{2}} |\sin x \cos \frac{π}{4} - \cos x \sin \frac{π}{4}| d x = \sqrt{2} \int_{0}^{\frac{π}{2}} |\sin (x - \frac{π}{4})| d x We have, |\sin (x - \frac{π}{4})| = \{\begin{cases} - \sin (x - \frac{π}{4}), 0 \leq x \leq \frac{π}{4} \\ \sin (x - \frac{π}{4}), \frac{π}{4} \leq x \leq \frac{π}{2} \end{cases} ∴ \int_{0}^{\frac{π}{2}} |\sin x - \cos x| d x = \sqrt{2} \int_{0}^{\frac{π}{4}} - \sin (x - \frac{π}{4}) d x + \sqrt{2} \int_{\frac{π}{4}}^{\frac{π}{2}} \sin (x - \frac{π}{4}) d x = \sqrt{2} {[\cos (x - \frac{π}{4})]}_{0}^{\frac{π}{4}} - \sqrt{2} {[\cos (x - \frac{π}{4})]}_{\frac{π}{4}}^{\frac{π}{2}} = \sqrt{2} [\cos (0) - \cos (- \frac{π}{4})] - \sqrt{2} [\cos (\frac{π}{4}) - \cos (0)] = \sqrt{2} (1 - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + 1) = \sqrt{2} (2 - \frac{2}{\sqrt{2}}) = 2 \sqrt{2} - 2 = 2 (\sqrt{2} - 1)$
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Page No 20.122:

Question 32:

$\int_{0}^{1} |\sin 2 π x| d x$

Answer:

$We have, |\sin 2 πx| = \{\begin{cases} (\sin 2 πx), 0 \leq x \leq \frac{1}{2} \\ - (\sin 2 πx), \frac{1}{2} \leq x \leq 1 \end{cases} ∴ \int_{0}^{1} |\sin 2 πx| d x = \int_{0}^{\frac{1}{2}} \sin 2 πx dx + \int_{\frac{1}{2}}^{1} - \sin 2 πx dx = {[\frac{- \cos 2 πx}{2 π}]}_{0}^{\frac{1}{2}} + {[\frac{\cos 2 πx}{2 π}]}_{\frac{1}{2}}^{1} = \frac{1}{2 π} + \frac{1}{2 π} + \frac{1}{2 π} + \frac{1}{2 π} = \frac{2}{π}$

Page No 20.122:

Question 33:

$\int_{1}^{3} |x^{2} - 4| d x$

Answer:

$\int_{1}^{3} |x^{2} - 4| d x = \int_{1}^{2} - (x^{2} - 4) d x + \int_{2}^{3} (x^{2} - 4) d x = {[- \frac{x^{3}}{3} + 4 x]}_{1}^{2} + {[\frac{x^{3}}{3} - 4 x]}_{2}^{3} = \frac{- 8}{3} + 8 + \frac{1}{3} - 4 + 9 - 12 - \frac{8}{3} + 8 = 4$

Page No 20.122:

Question 34:

$\int_{- π / 2}^{π / 2} \sin^{9} x d x$

Answer:

$\int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{9} x d x Let f (x) = \sin^{9} x Consider, f (- x) = \sin^{9} (- x) = - \sin^{9} x = - f (x) Thus f (x) is an odd function Therefore, \int_{\frac{- π}{2}}^{\frac{π}{2}} \sin^{9} x d x = 0$

Page No 20.122:

Question 35:

$\int_{- 1 / 2}^{1 / 2} \cos x \log (\frac{1 + x}{1 - x}) d x$

Answer:

$\int_{\frac{- 1}{2}}^{\frac{1}{2}} \cos x \log (\frac{1 + x}{1 - x}) d x Let f (x) = \cos x \log (\frac{1 + x}{1 - x}) Consider f (- x) = \cos (- x) \log (\frac{1 - x}{1 + x}) = \cos x \{- \log (\frac{1 + x}{1 - x})\} = - \cos x \log (\frac{1 + x}{1 - x}) = - f (x) Thus f (x) is an odd function Therefore, \int_{\frac{- 1}{2}}^{\frac{1}{2}} \cos x \log (\frac{1 + x}{1 - x}) d x = 0$

Page No 20.122:

Question 36:

$\int_{- a}^{a} \frac{x e^{x^{2}}}{1 + x^{2}} d x$

Answer:

$\int_{- a}^{a} \frac{x e^{x^{2}}}{1 + x^{2}} d x Let f (x) = \frac{x e^{x^{2}}}{1 + x^{2}} Consider f (- x) = - \frac{x e^{x^{2}}}{1 + x^{2}} = - f (x) Thus f (x) i s an odd function Therefore, \int_{- a}^{a} \frac{x e^{x^{2}}}{1 + x^{2}} d x = 0$ ode is 4430. →

Page No 20.122:

Question 37:

$\int_{0}^{π / 2} \frac{1}{1 + \cot^{7} x} d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t^{7} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t^{7} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{7} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t^{7} x} + \frac{1}{1 + \tan^{7} x} d x = \int_{0}^{\frac{π}{2}} \frac{2 + c o t^{7} x + \tan^{7} x}{(1 + c o t^{7} x) (1 + \tan^{7} x)} d x = \int_{0}^{\frac{π}{2}} \frac{2 + c o t^{7} x + \tan^{7} x}{2 + c o t^{7} x + \tan^{7} x} d x = \int_{0}^{\frac{π}{2}} d x = {[x]}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence, I = \frac{π}{4}$

Page No 20.122:

Question 38:

$\int_{0}^{2 π} \cos^{7} x d x$

Answer:

$Let, I = \int_{0}^{2 π} \cos^{7} x d x . . . (i) = \int_{0}^{2 π} \cos^{7} (2 π - x) d x = \int_{0}^{2 π} - \cos^{7} x d x \Rightarrow I = - \int_{0}^{2 π} \cos^{7} x d x . . . (ii) Adding (i) and (ii) we get, 2 I = \int_{0}^{2 π} \cos^{7} x d x - \int_{0}^{2 π} \cos^{7} x d x \Rightarrow 2 I = 0 ∴ I = 0$

Page No 20.122:

Question 39:

$\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x$

Answer:

$Let, I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x . . . (i) = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{a - a + x}} d x = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} d x \Rightarrow I = \int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{a} [\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} + \frac{\sqrt{a - x}}{\sqrt{x} + \sqrt{a - x}}] d x = \int_{0}^{a} d x = {[x]}_{0}^{a} = a Hence, I = \frac{a}{2}$

Page No 20.122:

Question 40:

$\int_{0}^{π / 2} \frac{1}{1 + \tan^{3} x} d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{1}{1 + \tan^{3} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{1}{1 + c o t^{3} x} d x . . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} [\frac{1}{1 + \tan^{3} x} + \frac{1}{1 + c o t^{3} x}] d x = \int_{0}^{\frac{π}{2}} \frac{2 + \tan^{3} x + c o t^{3} x}{(1 + \tan^{3} x) (1 + c o t^{3} x)} d x = \int_{0}^{\frac{π}{2}} \frac{2 + \tan^{3} x + c o t^{3} x}{2 + \tan^{3} x + c o t^{3} x} d x = \int_{0}^{\frac{π}{2}} d x = {(x)}_{0}^{\frac{π}{2}} = \frac{π}{2} Hence, I = \frac{π}{4}$

Page No 20.122:

Question 41:

$\int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x$

Answer:

$Let, I = \int_{0}^{π} \frac{x \sin x}{1 + \cos^{2} x} d x . . . (i) = \int_{0}^{π} \frac{(π - x) \sin (π - x)}{1 + \cos^{2} (π - x)} d x = \int_{0}^{π} \frac{(π - x) \sin x}{1 + \cos^{2} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{π} [\frac{x \sin x}{1 + \cos^{2} x} + \frac{(π - x) \sin x}{1 + \cos^{2} x}] d x = \int_{0}^{π} \frac{π \sin x}{1 + \cos^{2} x} d x = π {[- \tan^{- 1} (cosx)]}_{0}^{π} = - π [\tan^{- 1} (- 1) - \tan^{- 1} (1)] = - π (- \frac{π}{4} - \frac{π}{4}) = \frac{π^{2}}{2} Hence, I = \frac{π^{2}}{4}$

Page No 20.122:

Question 42:

$\int_{0}^{π} x \sin x \cos^{4} x d x$

Answer:

$Le t, I = \int_{0}^{π} x \sin x \cos^{4} x d x . . . (i) = \int_{0}^{π} (π - x) \sin (π - x) \cos^{4} (π - x) d x = \int_{0}^{π} (π - x) \sin x \cos^{4} x d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{π} [x \sin x \cos^{4} x + (π - x) \sin x \cos^{4} x] d x = \int_{0}^{π} (x + π - x) \sin x \cos^{4} x d x = π \int_{0}^{π} \sin x \cos^{4} x d x = π {[\frac{- \cos^{5} x}{5}]}_{0}^{π} = π [\frac{1}{5} + \frac{1}{5}] = \frac{2 π}{5} Hence, I = \frac{π}{5}$

Page No 20.122:

Question 43:

$\int_{0}^{π} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x$

Answer:

$We have, I = \int_{0}^{π} \frac{x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x . . . . . (1) = \int_{0}^{π} \frac{(π - x)}{a^{2} \cos^{2} (π - x) + b^{2} \sin^{2} (π - x)} d x = \int_{0}^{π} \frac{π - x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x . . . . . (2) Adding (1) and (2) 2 I = \int_{0}^{π} \frac{x + π - x}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x = π \int_{0}^{π} \frac{1}{a^{2} \cos^{2} x + b^{2} \sin^{2} x} d x = π \int_{0}^{π} \frac{\sec^{2} x}{a^{2} + b^{2} \tan^{2} x} dx (Dividing numerator and denominator by \cos^{2} x) = 2 π \int_{0}^{\frac{π}{2}} \frac{\sec^{2} x}{a^{2} + b^{2} \tan^{2} x} dx [Using \int_{0}^{2 a} f (x) d x = \int_{0}^{a} f (x) d x + \int_{0}^{a} f (2 a - x) d x] Putting \tan x = t \Rightarrow \sec^{2} x d x = d t When x \to 0; t \to 0 and x \to \frac{π}{2}; t \to \infty ∴ 2 I = 2 π \int_{0}^{\frac{π}{2}} \frac{d t}{a^{2} + b^{2} t^{2}} \Rightarrow I = \frac{π}{b^{2}} \int_{0}^{\frac{π}{2}} \frac{d t}{\frac{a^{2}}{b^{2}} + t^{2}} = \frac{π}{b^{2}} \times \frac{b}{a} {[\tan^{- 1} (\frac{b t}{a})]}_{0}^{\infty} = \frac{π}{a b} [\frac{π}{2} - 0] = \frac{π}{a b} \times \frac{π}{2} = \frac{π^{2}}{2 a b} Hence I = \frac{π^{2}}{2 a b}$

Page No 20.122:

Question 44:

$\int_{- π / 4}^{π / 4} |\tan x| d x$

Answer:

$\int_{\frac{- π}{4}}^{\frac{π}{4}} |\tan x| d x = \int_{\frac{- π}{4}}^{0} - \tan x d x + \int_{0}^{\frac{π}{4}} \tan x d x = {[\log (\cos x)]}_{\frac{- π}{4}}^{0} + {[- \log (\cos x)]}_{0}^{\frac{π}{4}} = - \log \frac{1}{\sqrt{2}} - \log \frac{1}{\sqrt{2}} = 2 \log \sqrt{2} = \log 2$

Page No 20.122:

Question 45:

$\int_{0}^{15} [x^{2}] d x$

Answer:

$We have, I = \int_{0}^{1.5} [x^{2}] d x = \int_{0}^{1} [x^{2}] d x + \int_{1}^{\sqrt{2}} [x^{2}] d x + \int_{\sqrt{2}}^{1.5} [x^{2}] d x = \int_{0}^{1} (0) d x + \int_{1}^{\sqrt{2}} (1) d x + \int_{\sqrt{2}}^{1.5} (2) d x (∵ [x^{2}] = \{\begin{cases} \begin{array}{l} 0 where, 0 < x < 1 \\ 1 where, 1 < x < \sqrt{2} \end{array} \\ 2 where, \sqrt{2} < x < 1.5 \end{cases}) = 0 + {[x]}_{1}^{\sqrt{2}} + {[2 x]}_{\sqrt{2}}^{1.5} = {[x]}_{1}^{\sqrt{2}} + 2 {[x]}_{\sqrt{2}}^{1.5} = (\sqrt{2} - 1) + 2 (1.5 - \sqrt{2}) = \sqrt{2} - 1 + 3 - 2 \sqrt{2} = 2 - \sqrt{2}$

Page No 20.122:

Question 46:

$\int_{0}^{π} \frac{x}{1 + \cos α \sin x} d x$

Answer:

$We have, I = \int_{0}^{π} \frac{x}{1 + \cos α \sin x} d x . . . . . (1) \Rightarrow I = \int_{0}^{π} \frac{π - x}{1 + \cos α \sin (π - x)} d x (∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x) \Rightarrow I = \int_{0}^{π} \frac{π - x}{1 + \cos α \sin x} d x . . . . . (2) Adding (1) and (2), we get$

$2 I = \int_{0}^{π} \frac{π}{1 + \cos α \sin x} d x \Rightarrow I = \frac{π}{2} \int_{0}^{π} \frac{1}{1 + \cos α \sin x} d x = \frac{π}{2} \int_{0}^{π} \frac{1}{1 + \cos α \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} d x = \frac{π}{2} \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} + \cos α 2 \tan \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t When x \to 0; t \to 0 and x \to π; t \to \infty Now, integral becomes$

$I = π \int_{0}^{\infty} \frac{d t}{1 + t^{2} + 2 t \cos α} = π \int_{0}^{\infty} \frac{d t}{{(t + \cos α)}^{2} + 1 - \cos^{2} α} = π \int_{0}^{\infty} \frac{d t}{{(t + \cos α)}^{2} + \sin^{2} α} = π {[\frac{1}{\sin α} \tan^{- 1} \frac{t + \cos α}{\sin α}]}_{0}^{\infty} = \frac{π}{\sin α} {[\tan^{- 1} \frac{t + \cos α}{\sin α}]}_{0}^{\infty} = \frac{π}{\sin α} [\frac{π}{2} - \tan^{- 1} (\cot α)] = \frac{π}{\sin α} [\frac{π}{2} - \tan^{- 1} \{\tan (\frac{π}{2} - α)\}] = \frac{π}{\sin α} [\frac{π}{2} - (\frac{π}{2} - α)] = \frac{π α}{\sin α}$

Page No 20.122:

Question 47:

$\int_{0}^{π / 2} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \frac{x \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x . . . (i) = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \sin (\frac{π}{2} - x) \cos (\frac{π}{2} - x)}{\sin^{4} (\frac{π}{2} - x) + \cos^{4} (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{(\frac{π}{2} - x) \cos x \sin x}{\cos^{4} x + \sin^{4} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{\frac{π}{2}} \frac{(x + \frac{π}{2} - x) \sin x \cos x}{\sin^{4} x + \cos^{4} x} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 - 2 \sin^{2} x \cos^{2} x} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 - 2 \sin^{2} x (1 - \sin^{2} x)} d x = \frac{π}{2} \int_{0}^{\frac{π}{2}} \frac{\sin x \cos x}{1 - 2 \sin^{2} x + 2 \sin^{4} x} d x Let, \sin^{2} x = t, then 2 \sin x \cos x d x = d t When, x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1 2 I = \frac{π}{4} \int_{0}^{1} \frac{1}{1 - 2 t + 2 t^{2}} d t = \frac{π}{8} \int_{0}^{1} \frac{1}{{(t - \frac{1}{2})}^{2} + \frac{1}{4}} = \frac{π}{8} {[2 \tan^{- 1} (2 t - 1)]}_{0}^{1} = \frac{π}{4} [\tan^{- 1} (1) - \tan^{- 1} (- 1)] = \frac{π}{4} [\frac{π}{4} + \frac{π}{4}] = \frac{π^{2}}{8} Hence, I = \frac{π^{2}}{16}$

Page No 20.122:

Question 48:

$\int_{0}^{π / 2} \frac{\cos^{2} x}{\sin x + \cos x} d x$

Answer:

$We have, I = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\sin x + \cos x} d x . . . . . (1) = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\cos x + \sin x} d x . . . . . (2)$

$Adding (1) and (2) 2 I = \int_{0}^{\frac{π}{2}} [\frac{\cos^{2} x}{\sin x + \cos x} + \frac{\sin^{2} x}{\cos x + \sin x}] d x = \int_{0}^{\frac{π}{2}} [\frac{1}{\sin x + \cos x}] d x = \int_{0}^{\frac{π}{2}} [\frac{1}{\frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}}] d x$

$= - \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{\tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2} - 1} d x = - \int_{0}^{\frac{π}{2}} \frac{\sec^{2} \frac{x}{2}}{\tan^{2} \frac{x}{2} - 2 \tan \frac{x}{2} - 1} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t When x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1$

$∴ 2 I = - 2 \int_{0}^{1} \frac{d t}{t^{2} - 2 t - 1} \Rightarrow I = - \int_{0}^{1} \frac{d t}{{(t - 1)}^{2} - {(\sqrt{2})}^{2}} = - \frac{1}{2 \sqrt{2}} {[\log |\frac{t - 1 - \sqrt{2}}{t - 1 + \sqrt{2}}|]}_{0}^{1} = - \frac{1}{2 \sqrt{2}} [\log |- 1| - \log |\frac{- 1 - \sqrt{2}}{- 1 + \sqrt{2}}|] = - \frac{1}{2 \sqrt{2}} [\log 1 - \log \frac{\sqrt{2} + 1}{\sqrt{2} - 1}]$

$= - \frac{1}{2 \sqrt{2}} [- \log \frac{\sqrt{2} + 1}{\sqrt{2} - 1}] = \frac{1}{2 \sqrt{2}} \log [\frac{(\sqrt{2} + 1) (\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)}] = \frac{1}{2 \sqrt{2}} \log [\frac{{(\sqrt{2} + 1)}^{2}}{(2 - 1)}] = \frac{1}{2 \sqrt{2}} \log {(\sqrt{2} + 1)}^{2} = \frac{1}{2 \sqrt{2}} \times 2 \log (\sqrt{2} + 1) = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

Page No 20.122:

Question 49:

$\int_{0}^{π} \cos 2 x \log \sin x d x$

Answer:

$\int_{0}^{π} \cos 2 x \log \sin x d x = {[\log \sin x \frac{\sin 2 x}{2}]}_{0}^{π} - \int_{0}^{π} \frac{\cos x}{\sin x} \frac{\sin 2 x}{2} d x = {[\log \sin x \frac{\sin 2 x}{2}]}_{0}^{π} - \int_{0}^{π} \cos^{2} x d x = {[\log \sin x \frac{\sin 2 x}{2}]}_{0}^{π} - \int_{0}^{π} \frac{1 + \cos 2 x}{2} d x = {[\log \sin x \frac{\sin 2 x}{2}]}_{0}^{π} - \frac{1}{2} {[x + \frac{\sin 2 x}{2}]}_{0}^{π} = 0 - \frac{1}{2} (π + 0) = - \frac{π}{2}$

Page No 20.122:

Question 50:

$\int_{0}^{π} \frac{x}{a^{2} - \cos^{2} x} d x, a > 1$

Answer:

$Let I = \int_{0}^{π} \frac{x}{a^{2} - \cos^{2} x} d x . . . (i) = \int_{0}^{π} \frac{π - x}{a^{2} - \cos^{2} (π - x)} d x = \int_{0}^{π} \frac{π - x}{a^{2} - \cos^{2} x} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{0}^{π} \frac{π}{a^{2} - \cos^{2} x} d x = \frac{π}{2 a} \int_{0}^{π} [\frac{1}{a - cosx} + \frac{1}{a + cosx}] dx = \frac{π}{2 a} \int_{0}^{π} [\frac{\sec^{2} \frac{x}{2}}{(a - 1) + (a + 1) \tan^{2} \frac{x}{2}} + \frac{\sec^{2} \frac{x}{2}}{(a + 1) + (a - 1) \tan^{2} \frac{x}{2}}] dx Let, \tan \frac{x}{2} = t, then \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt 2 I = \frac{π}{a} \int_{0}^{\infty} [\frac{1}{(a - 1) + (a + 1) t^{2}} + \frac{1}{(a + 1) + (a - 1) t^{2}}] dt = \frac{π}{a \sqrt{(a^{2} - 1)}} {[\tan^{- 1} \sqrt{\frac{a + 1}{a - 1}} t + \tan^{- 1} \sqrt{\frac{a - 1}{a + 1}} t]}_{0}^{\infty} = \frac{π}{a \sqrt{(a^{2} - 1)}} [\frac{π}{2} + \frac{π}{2}] = \frac{π^{2}}{a \sqrt{(a^{2} - 1)}} ∴ I = \frac{π^{2}}{2 a \sqrt{(a^{2} - 1)}}$

Page No 20.122:

Question 51:

$\int_{0}^{π} \frac{x \tan x}{\sec x + \tan x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{x \tan x}{s e c x + \tan x} d x . . . (i) = \int_{0}^{π} \frac{(π - x) \tan (π - x)}{s e c (π - x) + \tan (π - x)} d x = \int_{0}^{π} \frac{(π - x) \tan x}{s e c x + \tan x} d x . . . (ii) Adding (i) and (ii) we get 2 I = \int_{0}^{π} \frac{π \tan x}{s e c x + \tan x} d x = π \int_{0}^{π} \frac{sinx}{1 + sinx} dx = π \int_{0}^{π} \frac{1 + sinx - 1}{1 + sinx} dx = π \int_{0}^{π} [1 - \frac{1}{1 + sinx}] dx = π {[x]}_{0}^{π} - π \int_{0}^{π} \frac{1}{1 + \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}} dx = π^{2} - π \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2} + 2 \tan \frac{x}{2}} dx = π^{2} - π \int_{0}^{π} \frac{\sec^{2} \frac{x}{2}}{{(1 + \tan \frac{x}{2})}^{2}} dx = π^{2} + π {[\frac{2}{1 + \tan \frac{x}{2}}]}_{0}^{π} = π^{2} + π (0 - 2) = π^{2} - 2 π = π (π - 2) Hence I = \frac{π}{2} (π - 2)$

Page No 20.122:

Question 52:

$\int_{2}^{3} \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} d x$

Answer:

$Let, I = \int_{2}^{3} \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} d x . . . (i) = \int_{2}^{3} \frac{\sqrt{5 - x}}{\sqrt{5 - 5 + x} + \sqrt{5 - x}} d x = \int_{2}^{3} \frac{\sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}} d x . . . (ii) Adding (i) and (ii) 2 I = \int_{2}^{3} [\frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} + \frac{\sqrt{5 - x}}{\sqrt{x} + \sqrt{5 - x}}] d x = \int_{2}^{3} \frac{\sqrt{5 - x} + \sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} d x = \int_{2}^{3} d x = {[x]}_{2}^{3} = 3 - 1 = 1 Hence, I = \frac{1}{2}$

Page No 20.122:

Question 53:

$\int_{0}^{π / 2} \frac{\sin^{2} x}{\sin x + \cos x} d x$

Answer:

$We have, I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{\sin x + \cos x} d x . . . . . (1) = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} d x = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\cos x + \sin x} d x . . . . . (2) Adding (1) and (2) 2 I = \int_{0}^{\frac{π}{2}} [\frac{\sin^{2} x}{\sin x + \cos x} + \frac{\cos^{2} x}{\cos x + \sin x}] d x = \int_{0}^{\frac{π}{2}} [\frac{1}{\sin x + \cos x}] d x = \int_{0}^{\frac{π}{2}} [\frac{1 + \tan^{2} \frac{x}{2}}{2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}}] d x = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} \frac{x}{2}}{2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} d x = d t \Rightarrow \sec^{2} \frac{x}{2} d x = 2 d t When x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1 ∴ 2 I = \int_{0}^{1} \frac{2 d t}{2 t + 1 - t^{2}} d x = 2 \int_{0}^{1} \frac{d t}{{(\sqrt{2})}^{2} - {(t - 1)}^{2}} = \frac{2}{2 \sqrt{2}} {[\log |\frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1}|]}_{0}^{1} = \frac{1}{\sqrt{2}} [\log (\frac{\sqrt{2}}{\sqrt{2}}) - l o g |\frac{\sqrt{2} - 1}{\sqrt{2} + 1}|] = \frac{1}{\sqrt{2}} [0 - \log |\frac{\sqrt{2} - 1}{\sqrt{2} + 1}|] = - \frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} - 1}{\sqrt{2} + 1}| = \frac{1}{\sqrt{2}} \log |\frac{\sqrt{2} + 1}{\sqrt{2} - 1}| = \frac{1}{\sqrt{2}} \log [\frac{(\sqrt{2} + 1) (\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)}] 2 I = \frac{1}{\sqrt{2}} \log [\frac{{(\sqrt{2} + 1)}^{2}}{2 - 1}] 2 I = \frac{2}{\sqrt{2}} \log (\sqrt{2} + 1) Hence I = \frac{1}{\sqrt{2}} \log (\sqrt{2} + 1)$

Page No 20.122:

Question 54:

$\int_{0}^{π / 2} \frac{x}{\sin^{2} x + \cos^{2} x} d x$

Answer:

$Let, I = \int_{0}^{\frac{π}{2}} \frac{x}{\sin^{2} x + \cos^{2} x} d x = \int_{0}^{\frac{π}{2}} \frac{x}{1} d x = \int_{0}^{\frac{π}{2}} x d x = {[\frac{x^{2}}{2}]}_{0}^{\frac{π}{2}} ∴ I = \frac{π^{2}}{8}$

Page No 20.122:

Question 55:

$\int_{- π}^{π} x^{10} \sin^{7} x d x$

Answer:

$\int_{- π}^{π} x^{10} \sin^{7} x d x Let f (x) = x^{10} \sin^{7} x Consider f (- x) = {(- x)}^{10} \sin^{7} (- x) = - x^{10} \sin^{7} x = - f (x) Hence f (x) is an odd function Therefore \int_{- π}^{π} x^{10} \sin^{7} x d x = 0$

Page No 20.122:

Question 56:

$\int_{0}^{1} \cot^{- 1} (1 - x + x^{2}) d x$

Answer:

$\int_{0}^{1} c o t^{- 1} (1 - x + x^{2}) d x = \int_{0}^{1} c o t^{- 1} [x (x - 1) + 1] d x = \int_{0}^{1} c o t^{- 1} [\frac{(x (x - 1) + 1)}{x - (x - 1)}] d x = \int_{0}^{1} c o t^{- 1} x - c o t^{- 1} (x - 1) d x = {[x c o t^{- 1} x]}_{0}^{1} + \int_{0}^{1} \frac{x}{1 + x^{2}} d x - {[(x - 1) c o t^{- 1} (x - 1)]}_{0}^{1} - \int_{0}^{1} \frac{(x - 1)}{1 + {(x - 1)}^{2}} d x = {[x c o t^{- 1} x]}_{0}^{1} + \frac{1}{2} {[\log (1 + x^{2})]}_{0}^{1} - {[(x - 1) c o t^{- 1} (x - 1)]}_{0}^{1} - \frac{1}{2} {[\log (1 + {(1 - x)}^{2})]}_{0}^{1} = \frac{π}{4} - \frac{1}{2} \log 2 + \frac{π}{4} - \frac{1}{2} \log 2 = \frac{π}{2} - \log 2$

Page No 20.122:

Question 57:

$\int_{0}^{π} \frac{d x}{6 - \cos x}$

Answer:

$\int_{0}^{π} \frac{1}{6 - \cos x} d x = \int_{0}^{π} \frac{1 + \tan^{2} \frac{x}{2}}{6 + 6 \tan^{2} \frac{x}{2} - 1 + \tan^{2} \frac{x}{2}} d x = \int_{0}^{π} \frac{s e c^{2} \frac{x}{2}}{5 + 7 \tan^{2} \frac{x}{2}} d x Let, \tan \frac{x}{2} = t, then \frac{1}{2} s e c^{2} \frac{x}{2} d x = d t Therefore the integralbecomes \int_{0}^{\infty} \frac{2 d t}{5 + 7 t^{2}} = \frac{2}{7} \int_{0}^{\infty} \frac{d t}{\frac{5}{7} + t^{2}} = \frac{2}{\sqrt{35}} {[\tan^{- 1} \frac{\sqrt{7} t}{\sqrt{5}}]}_{0}^{\infty} = \frac{π}{\sqrt{35}}$

Page No 20.122:

Question 58:

$\int_{0}^{π / 2} \frac{1}{2 \cos x + 4 \sin x} d x$

Answer:

$We have, I = \int_{0}^{\frac{π}{2}} \frac{1}{2 \cos x + 4 \sin x} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{2 - 2 \tan^{2} \frac{x}{2} + 8 \tan \frac{x}{2}} d x Putting \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} s e c^{2} \frac{x}{2} d x = d t When x \to 0; t \to 0 and x \to \frac{π}{2}; t \to 1 ∴ I = 2 \int_{0}^{1} \frac{d t}{2 - 2 t^{2} + 8 t} = - \frac{2}{2} \int_{0}^{1} \frac{d t}{t^{2} - 4 t - 1} = - \int_{0}^{1} \frac{d t}{{(t - 2)}^{2} - 5} = \int_{0}^{1} \frac{d t}{{(\sqrt{5})}^{2} - {(t - 2)}^{2}} = \frac{1}{2 \sqrt{5}} {[\log |\frac{\sqrt{5} + t - 2}{\sqrt{5} - t + 2}|]}_{0}^{1} = \frac{1}{2 \sqrt{5}} [\log \frac{\sqrt{5} - 1}{\sqrt{5} + 1} - \log \frac{\sqrt{5} - 2}{\sqrt{5} + 2}] = \frac{1}{2 \sqrt{5}} \log [\frac{\sqrt{5} - 1}{\sqrt{5} + 1} \times \frac{\sqrt{5} + 2}{\sqrt{5} - 2}] = \frac{1}{2 \sqrt{5}} \log [\frac{5 + 2 \sqrt{5} - \sqrt{5} - 2}{5 - 2 \sqrt{5} + \sqrt{5} - 2}] = \frac{1}{2 \sqrt{5}} \log [\frac{\sqrt{5} + 3}{- \sqrt{5} + 3}]$
$I = \frac{1}{2 \sqrt{5}} \log (\frac{3 + \sqrt{5}}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}}) I = \frac{1}{2 \sqrt{5}} l o g {(\frac{3 + \sqrt{5}}{2})}^{2} I = \frac{2}{2 \sqrt{5}} l o g (\frac{3 + \sqrt{5}}{2}) I = \frac{1}{\sqrt{5}} l o g (\frac{3 + \sqrt{5}}{2})$

Page No 20.122:

Question 59:

$\int_{π / 6}^{π / 2} \frac{cosec x \cot x}{1 + {cosec}^{2} x} d x$

Answer:

$\int_{\frac{π}{6}}^{\frac{π}{2}} \frac{\cos e c x c o t x}{1 + \cos e c^{2} x} d x = \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{\cos x}{1 + \sin^{2} x} d x = {[\tan^{- 1} (\sin x)]}_{\frac{π}{6}}^{\frac{π}{2}} = \tan^{- 1} 1 - \tan^{- 1} \frac{1}{2} = \tan^{- 1} \frac{1 - \frac{1}{2}}{1 + 1 \times \frac{1}{2}} = \tan^{- 1} \frac{1}{3}$

Page No 20.122:

Question 60:

$\int_{0}^{π / 2} \frac{d x}{4 \cos x + 2 \sin x}$

Answer:

$\int_{0}^{\frac{π}{2}} \frac{1}{4 \cos x + 2 \sin x} d x = \int_{0}^{\frac{π}{2}} \frac{1 + \tan^{2} \frac{x}{2}}{4 - 4 \tan^{2} \frac{x}{2} + 4 \tan \frac{x}{2}} d x Let \tan \frac{x}{2} = t, then \frac{1}{2} s e c^{2} \frac{x}{2} d x = d t When x = 0, t = 0, x = \frac{π}{2}, t = 1 = \frac{- 1}{4} \int_{0}^{1} \frac{d t}{{(t - \frac{1}{2})}^{2} - \frac{5}{4}} = \frac{- 1}{4} \times \frac{- 4}{\sqrt{5}} {[\log \frac{2 t - 1 - \sqrt{5}}{2 t - 1 + \sqrt{5}}]}_{0}^{1} = \frac{1}{\sqrt{5}} \log \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$

Page No 20.122:

Question 61:

$\int_{0}^{4} x d x$

Answer:

$\int_{0}^{4} x d x = {[\frac{x^{2}}{2}]}_{0}^{4} = 8 - 0 = 8$

Page No 20.122:

Question 62:

$\int_{0}^{2} (2 x^{2} + 3) d x$

Answer:

$\int_{0}^{2} (2 x^{2} + 3) d x = {[\frac{2 x^{3}}{3} + 3 x]}_{0}^{2} = \frac{16}{3} + 6 = \frac{34}{3}$

Page No 20.122:

Question 63:

$\int_{1}^{4} (x^{2} + x) d x$

Answer:

$Here a = 1, b = 4, f (x) = x^{2} + x, h = \frac{4 - 1}{n} = \frac{3}{n} Therefore, \int_{1}^{4} (x^{2} + x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [1 + 1 + {(1 + h)}^{2} + (1 + h) + {(1 + 2 h)}^{2} + (1 + 2 h) + . . . . . . . . . + {(1 + (n - 1) h)}^{2} + (1 + (n - 1) h)] = \lim_{h \to 0} h [2 n + h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2}) + 2 h (1 + 2 + . . . . . . + (n - 1)) + h (1 + 2 + . . . . . . + (n - 1))] = \lim_{h \to 0} h [2 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 3 h \frac{n (n - 1)}{2}] = \lim_{n \to 0 \infty} [6 + \frac{9}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + \frac{9}{2} (1 - \frac{1}{n})] = 6 + 9 + \frac{9}{2} = \frac{27}{2}$

Page No 20.122:

Question 64:

$\int_{- 1}^{1} e^{2 x} d x$

Answer:

$Here a = - 1, b = 1, f (x) = e^{2 x}, h = \frac{1 + 1}{n} = \frac{2}{n} Therefore, \int_{- 1}^{1} e^{2 x} d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (- 1) + f (- 1 + h) + . . . . . . . . . . + f (- 1 + (n - 1) h)] = \lim_{h \to 0} h [e^{- 2} + e^{2 (- 1 + h)} + e^{2 (- 1 + 2 h)} + . . . . . . . + e^{2 (- 1 + (n - 1) h)}] = \lim_{h \to 0} h e^{- 2} [\frac{{(e^{2 h})}^{n} - 1}{e^{2 h} - 1}] = \lim_{h \to 0} e^{- 2} [\frac{e^{4} - 1}{\frac{e^{2 h} - 1}{2 h}}] \times \frac{1}{2} (Since, n h = 2) = \frac{1}{2} (e^{2} - e^{- 2})$

Page No 20.122:

Question 65:

$\int_{2}^{3} e^{- x} d x$

Answer:

$Here a = 2, b = 3, f (x) = e^{- x}, h = \frac{3 - 2}{n} = \frac{1}{n} Therefore, \int_{2}^{3} e^{- x} d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (2) + f (2 + h) + . . . . . . . . . . + f (2 + (n - 1) h)] = \lim_{h \to 0} h [e^{- 2} + e^{- (2 + h)} + e^{- (2 + 2 h)} + . . . . . . . + e^{- (2 + (n - 1) h)}] = \lim_{h \to 0} h e^{- 2} [\frac{{(e^{- h})}^{n} - 1}{e^{- h} - 1}] = \lim_{h \to 0} e^{- 2} [\frac{e^{- 1} - 1}{\frac{e^{- h} - 1}{- h}}] \times - 1 (Since n h = 1) = (e^{- 2} - e^{- 3})$

Page No 20.122:

Question 66:

$\int_{1}^{3} (2 x^{2} + 5 x) d x$

Answer:

$Here, a = 1, b = 3, f (x) = 2 x^{2} + 5 x, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, \int_{1}^{3} (2 x^{2} + 5 x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [2 + 5 + 2 {(1 + h)}^{2} + 5 (1 + h) + 2 {(1 + 2 h)}^{2} + 5 (1 + 2 h) + . . . . . . . . . + 2 {((n - 1) h)}^{2} + 5 ((n - 1) h)] = \lim_{h \to 0} h [2 n + 2 h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2}) + 4 h (1 + 2 + . . . . . . . . . . . . (n - 1)) + 5 n + 5 h (1 + 2 + . . . . . . . . . . . . (n - 1))] = \lim_{h \to 0} h [7 n + 2 h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 9 h \frac{n (n - 1)}{2}] = \lim_{n \to 0 \infty} [14 + \frac{8}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + 18 (1 - \frac{1}{n})] = 14 + \frac{16}{3} + 18 = \frac{112}{3}$

Page No 20.122:

Question 67:

$\int_{1}^{3} (x^{2} + 3 x) d x$

Answer:

$Here a = 1, b = 3, f (x) = x^{2} + 3 x, h = \frac{3 - 1}{n} = \frac{2}{n} Therefore, \int_{1}^{3} (x^{2} + 3 x) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (1) + f (1 + h) + . . . . . . . . . . + f (1 + (n - 1) h)] = \lim_{h \to 0} h [1 + 3 + {(1 + h)}^{2} + 3 (1 + h) + {(1 + 2 h)}^{2} + 3 (1 + 2 h) + . . . . . . . . . + {((n - 1) h)}^{2} + 3 ((n - 1) h)] = \lim_{h \to 0} h [n + h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2}) + 2 h (1 + 2 + . . . . . . . . . . . . (n - 1)) + 3 n + 3 h (1 + 2 + . . . . . . . . . . . . (n - 1))] = \lim_{h \to 0} h [4 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6} + 5 h \frac{n (n - 1)}{2}] = \lim_{n \to 0 \infty} [8 + \frac{4}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n}) + 10 (1 - \frac{1}{n})] = 8 + \frac{8}{3} + 10 = \frac{62}{3}$

Page No 20.122:

Question 68:

$\int_{0}^{2} (x^{2} + 2) d x$

Answer:

$Here a = 0, b = 2, f (x) = x^{2} + 2, h = \frac{2 - 0}{n} = \frac{2}{n} Therefore, \int_{0}^{2} (x^{2} + 2) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [0 + 2 + {(0 + h)}^{2} + 2 + {(0 + 2 h)}^{2} + 2 + . . . . . . . . . + {((n - 1) h)}^{2} + 2] = \lim_{h \to 0} h [2 n + h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2})] = \lim_{h \to 0} h [2 n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{n \to 0 \infty} [4 + \frac{4}{3} (1 - \frac{1}{n}) (2 - \frac{1}{n})] = 4 + \frac{8}{3} = \frac{20}{3}$

Page No 20.122:

Question 69:

$\int_{0}^{3} (x^{2} + 1) d x$

Answer:

$Here, a = 0, b = 3, f (x) = x^{2} + 1, h = \frac{3 - 0}{n} = \frac{3}{n} Therefore, \int_{0}^{3} (x^{2} + 1) d x = \lim_{h \to 0} h [f (a) + f (a + h) + f (a + 2 h) + . . . . . . . . . . . . + f (a + (n - 1) h)] = \lim_{h \to 0} h [f (0) + f (0 + h) + . . . . . . . . . . + f (0 + (n - 1) h)] = \lim_{h \to 0} h [0 + 1 + h^{2} + 1 + {(2 h)}^{2} + 1 + . . . . . . . . . + {((n - 1) h)}^{2} + 1] = \lim_{h \to 0} h [n + h^{2} (1^{2} + 2^{2} + . . . . . . . . . . . . . . {(n - 1)}^{2})] = \lim_{h \to 0} h [n + h^{2} \frac{n (n - 1) (2 n - 1)}{6}] = \lim_{h \to 0} [3 + \frac{9}{2} (1 - \frac{1}{n}) (2 - \frac{1}{n})] = 3 + 9 = 12$

Page No 20.16:

Question 1:

$\int_{4}^{9} \frac{1}{\sqrt{x}} d x$

Answer:

$Let I = \int_{4}^{9} \frac{1}{\sqrt{x}} d x . Then, I = 2 \int_{4}^{9} \frac{1}{2 \sqrt{x}} d x \Rightarrow I = 2 {[\sqrt{x}]}_{4}^{9} \Rightarrow I = 2 (3 - 2) \Rightarrow I = 2$

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Question 2:

$\int_{- 2}^{3} \frac{1}{x + 7} d x$

Answer:

$Let I = \int_{- 2}^{3} \frac{1}{x + 7} d x . Then, I = {[\log (x + 7)]}_{- 2}^{3} \Rightarrow I = \log 10 - \log 5 \Rightarrow I = \log \frac{10}{5} [∵ \log a - \log b = \log \frac{a}{b}] \Rightarrow I = \log 2$

Page No 20.16:

Question 3:

$\int_{0}^{1 / 2} \frac{1}{\sqrt{1 - x^{2}}} d x$

Answer:

$Let I = \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - x^{2}}} d x . Then, I = {[\sin^{- 1} x]}_{0}^{\frac{1}{2}} \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} 0 \Rightarrow I = \frac{π}{6} - 0 \Rightarrow I = \frac{π}{6}$

Page No 20.16:

Question 4:

$\int_{0}^{1} \frac{1}{1 + x^{2}} d x$

Answer:

$Let I = \int_{0}^{1} \frac{1}{1 + x^{2}} d x . Then, I = {[\tan^{- 1} x]}_{0}^{1} \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} 0 \Rightarrow I = \frac{π}{4} - 0 \Rightarrow I = \frac{π}{4}$

Page No 20.16:

Question 5:

$\int_{2}^{3} \frac{x}{x^{2} + 1} d x$

Answer:

$Let I = \int_{2}^{3} \frac{x}{x^{2} + 1} d x . Then, I = \frac{1}{2} \int_{2}^{3} \frac{2 x}{x^{2} + 1} \Rightarrow I = \frac{1}{2} {[\log (x^{2} + 1)]}_{2}^{3} \Rightarrow I = \frac{1}{2} (\log 10 - \log 5) \Rightarrow I = \frac{1}{2} \log \frac{10}{5} [∵ \log a - \log b = \log \frac{a}{b}] \Rightarrow I = \frac{1}{2} \log 2$

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Question 6:

$\int_{0}^{\infty} \frac{1}{a^{2} + b^{2} x^{2}} d x$

Answer:

$Let I = \int_{0}^{\infty} \frac{1}{a^{2} + b^{2} x^{2}} d x . Then, I = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{1 + \frac{b^{2} x^{2}}{a^{2}}} d x \Rightarrow I = \frac{1}{a^{2}} \int_{0}^{\infty} \frac{1}{1 + {(\frac{b x}{a})}^{2}} d x \Rightarrow I = \frac{a}{b a^{2}} {[\tan^{- 1} (\frac{b x}{a})]}_{0}^{\infty} \Rightarrow I = \frac{1}{a b} (\tan^{- 1} \infty - \tan^{- 1} 0) \Rightarrow I = \frac{π}{2 a b}$

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Question 7:

$\int_{- 1}^{1} \frac{1}{1 + x^{2}} d x$

Answer:

$Let I = \int_{- 1}^{1} \frac{1}{1 + x^{2}} d x . Then, I = {[\tan^{- 1} x]}_{- 1}^{1} \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} (- 1) \Rightarrow I = \frac{π}{4} - (- \frac{π}{4}) \Rightarrow I = \frac{π}{2}$

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Question 8:

$\int_{0}^{\infty} e^{- x} d x$

Answer:

$Let I = \int_{0}^{\infty} e^{- x} d x . Then, I = {[- e^{- x}]}_{0}^{\infty} \Rightarrow I = - e^{- \infty} + e^{0} \Rightarrow I = 0 + 1 \Rightarrow I = 1$

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Question 9:

$\int_{0}^{1} \frac{x}{x + 1} d x$

Answer:

$Let I = \int_{0}^{1} \frac{x}{x + 1} d x . Then, I = \int_{0}^{1} 1 - \frac{1}{x + 1} d x \Rightarrow I = {[x - \log (x + 1)]}_{0}^{1} \Rightarrow I = 1 - \log 2 - (0 - \log 1) \Rightarrow I = \log e - \log 2 \Rightarrow I = \log \frac{e}{2}$

Page No 20.16:

Question 10:

$\int_{0}^{π / 2} (\sin x + \cos x) d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} (\sin x + \cos x) d x . Then, I = {[- \cos x + \sin x]}_{0}^{\frac{π}{2}} \Rightarrow I = 0 + 1 - (- 1 + 0) \Rightarrow I = 2$

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Question 11:

$\int_{π / 4}^{π / 2} \cot x d x$

Answer:

$Let I = \int_{\frac{π}{4}}^{\frac{π}{2}} \cot x d x . Then, I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \cot x \frac{- (cosec x + \cot x)}{cosec x + \cot x} d x \Rightarrow I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- cosec x \cot x - \cot^{2} x}{cosec x + \cot x} d x \Rightarrow I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- cosec x \cot x - {cosec}^{2} x + 1}{cosec x + \cot x} d x [∵ {cosec}^{2} x = 1 + \cot^{2} x] \Rightarrow I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- cosec x \cot x - {cosec}^{2} x}{cosec x + \cot x} d x - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{1}{cosec x + \cot x} d x \Rightarrow I = - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{- cosec x \cot x - {cosec}^{2} x}{cosec x + \cot x} d x - \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\sin x}{1 + \cos x} d x \Rightarrow I = - {[\log (cosec x + \cot x)]}_{\frac{π}{4}}^{\frac{π}{2}} + {[\log (1 + \cos x)]}_{\frac{π}{4}}^{\frac{π}{2}} \Rightarrow I = - \log (1 + \infty) + \log (\sqrt{2} + 1) + \log (1 + 0) - \log (1 + \frac{1}{\sqrt{2}}) \Rightarrow I = \log (\sqrt{2} + 1) - \log (\frac{\sqrt{2} + 1}{\sqrt{2}}) \Rightarrow I = \log (\frac{\sqrt{2} (\sqrt{2} + 1)}{(\sqrt{2} + 1)}) \Rightarrow I = \log \sqrt{2} \Rightarrow I = \frac{1}{2} \log 2$

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Question 12:

$\int_{0}^{π / 4} \sec x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{4}} \sec x d x . Then, I = \int_{0}^{\frac{π}{4}} \sec x \frac{\sec x + \tan x}{\sec x + \tan x} d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{\sec^{2} x + \sec x \tan x}{\sec x + \tan x} d x Put u = \sec x + \tan x \Rightarrow d u = \sec^{2} x + \sec x \tan x d x ∴ \int_{0}^{\frac{π}{4}} \frac{\sec^{2} x + \sec x \tan x}{\sec x + \tan x} d x = \int \frac{d u}{u} \Rightarrow I = [\log u] \Rightarrow I = {[\log (\sec x + \tan x)]}_{0}^{\frac{π}{4}} \Rightarrow I = \log (\sec \frac{π}{4} + \tan \frac{π}{4}) - \log (\sec 0 + \tan 0) \Rightarrow I = \log (\sqrt{2} + 1) - \log 1 \Rightarrow I = \log (\sqrt{2} + 1)$

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Question 13:

$\int_{π / 6}^{π / 4} cosec x d x$

Answer:

$Let I = \int_{\frac{π}{6}}^{\frac{π}{4}} cosec x d x . Then, I = \int_{\frac{π}{6}}^{\frac{π}{4}} cosec x \frac{cosec x - \cot x}{cosec x - \cot x} d x \Rightarrow I = \int_{\frac{π}{6}}^{\frac{π}{4}} \frac{{cosec}^{2} x - cosec x \cot x}{cosec x - \cot x} d x \Rightarrow I = {[\log (cosec x - \cot x)]}_{\frac{π}{6}}^{\frac{π}{4}} \Rightarrow I = \log (\sqrt{2} - 1) - \log (2 - \sqrt{3})$

Page No 20.16:

Question 14:

$\int_{0}^{1} \frac{1 - x}{1 + x} d x$

Answer:

$Let I = \int_{0}^{1} \frac{1 - x}{1 + x} d x . Then, I = \int_{0}^{1} (\frac{1}{1 + x} - \frac{1 + x - 1}{1 + x}) d x I = \int_{0}^{1} (\frac{1}{1 + x} - 1 + \frac{x}{1 + x}) d x \Rightarrow I = {[\log (1 + x) - x + \log (1 + x)]}_{0}^{1} \Rightarrow I = (\log 2 - 1 + \log 2) - (\log 1 - 0 + \log 1) = 2 \log 2 - 1$

Page No 20.16:

Question 15:

$\int_{0}^{π} \frac{1}{1 + \sin x} d x$

Answer:

$Let I = \int_{0}^{π} \frac{1}{1 + \sin x} d x . Then, I = \int_{0}^{π} \frac{1 - \sin x}{(1 + \sin x) (1 - \sin x)} d x \Rightarrow I = \int_{0}^{π} \frac{1 - \sin x}{1 - \sin^{2} x} d x \Rightarrow I = \int_{0}^{π} \frac{1 - \sin x}{\cos^{2} x} d x [∵ \sin^{2} x + \cos^{2} x = 1] \Rightarrow I = \int_{0}^{π} \sec^{2} x - \sec x \tan x d x \Rightarrow I = {[\tan x - \sec x]}_{0}^{π} \Rightarrow I = (\tan π - \sec π) - (\tan 0 - \sec 0) \Rightarrow I = 0 + 1 - (0 - 1) \Rightarrow I = 1 + 1 \Rightarrow I = 2$

Page No 20.16:

Question 16:

$\int_{- π / 4}^{π / 4} \frac{1}{1 + \sin x} d x$

Answer:

$Let I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{1 + \sin x} d x . Then, I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1 - \sin x}{1 - \sin^{2} x} d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{1 - \sin x}{\cos^{2} x} d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\frac{1}{\cos^{2} x} - \frac{\sin x}{\cos^{2} x}) d x \Rightarrow I = \int_{- \frac{π}{4}}^{\frac{π}{4}} (\sec^{2} x - \sec x \tan x) d x \Rightarrow I = {[\tan x - \sec x]}_{- \frac{π}{4}}^{\frac{π}{4}} \Rightarrow I = (1 - \sqrt{2}) - (- 1 - \sqrt{2}) \Rightarrow I = 2$

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Question 17:

$\int_{0}^{π / 2} \cos^{2} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \cos^{2} x d x . Then, I = \int_{0}^{\frac{π}{2}} \cos^{2} x d x \Rightarrow I = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + \cos 2 x) d x [∵ \cos 2 x = 2 \cos^{2} x - 1] \Rightarrow I = {[\frac{x}{2} + \frac{\sin 2 x}{4}]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π}{4} + 0 - 0 \Rightarrow I = \frac{π}{4}$

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Question 18:

$\int_{0}^{π / 2} \cos^{3} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \cos^{3} x d x . Then, I = \int_{0}^{\frac{π}{2}} \cos^{2} x \cos x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (1 - \sin^{2} x) \cos x d x Let u = \sin x, d u = \cos x d x \Rightarrow I = \int (1 - u^{2}) d u \Rightarrow I = [u - \frac{u^{3}}{3}] \Rightarrow I = {[\sin x - \frac{\sin^{3} x}{3}]}_{0}^{\frac{π}{2}} \Rightarrow I = 1 - \frac{1}{3} - 0 \Rightarrow I = \frac{2}{3}$

Page No 20.16:

Question 19:

$\int_{0}^{π / 6} \cos x \cos 2 x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{6}} \cos x \cos 2 x d x . Then, I = \int_{0}^{\frac{π}{6}} \cos x (\cos^{2} x - \sin^{2} x) d x \Rightarrow I = \int_{0}^{\frac{π}{6}} (2 \cos^{3} x - \cos x) d x \Rightarrow I = \int_{0}^{\frac{π}{6}} (2 \cos x (1 - \sin^{2} x) - \cos x) d x \Rightarrow I = {[2 (\sin x - \frac{\sin^{3} x}{3}) - \sin x]}_{0}^{\frac{π}{6}} \Rightarrow I = [2 (\frac{1}{2} - \frac{1}{24}) - \frac{1}{2}] - 0 \Rightarrow I = \frac{5}{12}$

Page No 20.16:

Question 20:

$\int_{0}^{π / 2} \sin x \sin 2 x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sin x \sin 2 x d x . Then, I = \int_{0}^{\frac{π}{2}} 2 \sin^{2} x \cos x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} 2 (1 - \cos^{2} x) \cos x d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (2 \cos x - 2 \cos^{3} x) d x \Rightarrow I = {[2 \sin x - 2 (\sin x - \frac{\sin^{3} x}{3})]}_{0}^{\frac{π}{2}} \Rightarrow I = [2 - 2 (1 - \frac{1}{3})] - 0 \Rightarrow I = \frac{2}{3}$

Page No 20.16:

Question 21:

$\int_{π / 3}^{π / 4} {(\tan x + \cot x)}^{2} d x$

Answer:

$Let I = \int_{\frac{π}{3}}^{\frac{π}{4}} {(\tan x + \cot x)}^{2} d x . Then, I = \int_{\frac{π}{3}}^{\frac{π}{4}} (\tan^{2} x + \cot^{2} x + 2 \tan x \cot x) d x \Rightarrow I = \int_{\frac{π}{3}}^{\frac{π}{4}} (\tan^{2} x + \cot^{2} x + 2) d x \Rightarrow I = \int_{\frac{π}{3}}^{\frac{π}{4}} (\sec^{2} x - 1 + {cosec}^{2} x - 1 + 2) d x \Rightarrow I = \int_{\frac{π}{3}}^{\frac{π}{4}} (\sec^{2} x + {cosec}^{2} x) d x \Rightarrow I = {[\tan x - \cot x]}_{\frac{π}{3}}^{\frac{π}{4}} \Rightarrow I = (1 - 1) - (\sqrt{3} - \frac{1}{\sqrt{3}}) \Rightarrow I = \frac{- 2}{\sqrt{3}}$

Page No 20.16:

Question 22:

$\int_{0}^{π / 2} \cos^{4} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \cos^{4} x d x . Then, I = \int_{0}^{\frac{π}{2}} {(\cos^{2} x)}^{2} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{{(1 + \cos 2 x)}^{2}}{4} d x \Rightarrow I = \frac{1}{4} \int_{0}^{\frac{π}{2}} (1 + \cos^{2} 2 x + 2 \cos 2 x) d x \Rightarrow I = \frac{1}{4} \int_{0}^{\frac{π}{2}} (1 + 2 \cos 2 x + \frac{1 + \cos 4 x}{2}) d x \Rightarrow I = \frac{1}{4} \int_{0}^{\frac{π}{2}} (\frac{3 + 4 \cos 2 x + \cos 4 x}{2}) d x \Rightarrow I = \frac{1}{4} {[\frac{3 x}{2} + \frac{2 \sin 2 x}{2} + \frac{\sin 4 x}{8}]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{1}{4} [\frac{3 π}{4} + 0] \Rightarrow I = \frac{3 π}{16}$

Page No 20.16:

Question 23:

$\int_{0}^{π / 2} (a^{2} \cos^{2} x + b^{2} \sin^{2} x) d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} (a^{2} \cos^{2} x + b^{2} \sin^{2} x) d x . Then, I = \int_{0}^{\frac{π}{2}} (a^{2} \cos^{2} x + b^{2} (1 - \cos^{2} x)) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (b^{2} + (a^{2} - b^{2}) \cos^{2} x) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (b^{2} + \frac{(a^{2} - b^{2}) (1 + \cos 2 x)}{2}) d x \Rightarrow I = {[b^{2} x + \frac{a^{2} - b^{2}}{2} (x + \frac{\sin 2 x}{2})]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{b^{2} π}{2} + \frac{a^{2} - b^{2}}{2} \frac{π}{2} + 0 \Rightarrow I = \frac{π}{4} (a^{2} + b^{2})$

Page No 20.16:

Question 24:

$\int_{0}^{π / 2} \sqrt{1 + \sin x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sqrt{1 + \sin x} d x . Then, I = \int_{0}^{\frac{π}{2}} \sqrt{1 + \sin x} \times \frac{\sqrt{1 - \sin x}}{\sqrt{1 - \sin x}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{1 - \sin^{2} x}}{\sqrt{1 - \sin x}} d x \Rightarrow I = = \int_{0}^{\frac{π}{2}} \frac{\cos x}{\sqrt{1 - \sin x}} d x Let 1 - \sin x = u \Rightarrow - \cos x d x = d u ∴ I = \int \frac{- d u}{\sqrt{u}} \Rightarrow I = = [- 2 \sqrt{u}] \Rightarrow I = = {[- 2 \sqrt{1 - \sin x}]}_{0}^{\frac{π}{2}} \Rightarrow I = = 0 + 2 \Rightarrow I = = 2$

Page No 20.16:

Question 25:

$\int_{0}^{π / 2} \sqrt{1 + \cos x} d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} \sqrt{1 + \cos x} d x . Then, I = \int_{0}^{\frac{π}{2}} \sqrt{1 + \cos x} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sqrt{1 - \cos^{2} x}}{\sqrt{1 - \cos x}} d x \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin x}{\sqrt{1 - \cos x}} d x Let 1 - \cos x = u \Rightarrow \sin x d x = d u ∴ I = \int \frac{d u}{\sqrt{u}} \Rightarrow I = [2 \sqrt{u}] \Rightarrow I = {[2 \sqrt{1 - \cos x}]}_{0}^{\frac{π}{2}} \Rightarrow I = 2 - 0 \Rightarrow I = 2$

Page No 20.16:

Question 26:

Evaluate the following definite integrals:

$\int_{0}^{\frac{π}{2}} x^{2} \sin x d x$ [CBSE 2014]

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \sin x$

Applying integration by parts, we have

$I = x^{2} (- \cos x) |_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x (- \cos x) d x \Rightarrow I = (0 - 0) + 2 \int_{0}^{\frac{π}{2}} x \cos x d x (\cos \frac{π}{2} = 0)$

Again applying integration by parts, we have

$I = 0 + 2 [x \sin x |_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 \times \sin x d x] \Rightarrow I = 2 (\frac{π}{2} \sin \frac{π}{2} - 0) - 2 \int_{0}^{\frac{π}{2}} \sin x d x \Rightarrow I = 2 (\frac{π}{2} - 0) - 2 (- \cos x) |_{0}^{\frac{π}{2}} \Rightarrow I = π + 2 (\cos \frac{π}{2} - \cos 0) \Rightarrow I = π + 2 (0 - 1) \Rightarrow I = π - 2$

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Question 27:

$\int_{0}^{π / 2} x \cos x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x \cos x d x . Then, Integrating by parts I = {[x \sin x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 \sin x d x \Rightarrow I = {[x \sin x]}_{0}^{\frac{π}{2}} + {[\cos x]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π}{2} - 1$

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Question 28:

$\int_{0}^{π / 2} x^{2} \cos x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \cos x d x . Then, Integrating by parts I = {[x^{2} \sin x]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x \sin x d x \Rightarrow I = {[x^{2} \sin x]}_{0}^{\frac{π}{2}} - {[- 2 x \cos x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} - 2 \cos x d x \Rightarrow I = {[x^{2} \sin x]}_{0}^{\frac{π}{2}} + {[2 x \cos x]}_{0}^{\frac{π}{2}} - {[2 \sin x]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π^{2}}{4} - 2$

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Question 29:

$\int_{0}^{π / 4} x^{2} \sin x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{4}} x^{2} \sin x d x . Then, Integrating by parts I = {[- x^{2} \cos x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} - 2 x \cos x d x \Rightarrow I = {[- x^{2} \cos x]}_{0}^{\frac{π}{4}} + {[2 x \sin x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} 2 \sin x dx \Rightarrow I = {[- x^{2} \cos x]}_{0}^{\frac{π}{4}} + {[2 x \sin x]}_{0}^{\frac{π}{4}} + {[2 \cos x]}_{0}^{\frac{π}{4}} \Rightarrow I = \frac{- π^{2}}{16 \sqrt{2}} + \frac{π}{2 \sqrt{2}} + \frac{2}{\sqrt{2}} - 2 \Rightarrow I = \sqrt{2} + \frac{π}{2 \sqrt{2}} - \frac{π^{2}}{16 \sqrt{2}} - 2$

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Question 30:

$\int_{0}^{π / 2} x^{2} \cos 2 x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \cos 2 x d x . Then, Integrating by parts I = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 2 x \frac{\sin 2 x}{2} d x \Rightarrow I = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} - {[- x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} - 1 \frac{\cos 2 x}{2} d x \Rightarrow I = {[x^{2} \frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}} + {[x \frac{\cos 2 x}{2}]}_{0}^{\frac{π}{2}} - {[\frac{\sin 2 x}{4}]}_{0}^{\frac{π}{2}} \Rightarrow I = 0 - \frac{π}{4} - 0 \Rightarrow I = - \frac{π}{4}$

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Question 31:

$\int_{0}^{π / 2} x^{2} \cos^{2} x d x$

Answer:

$Let I = \int_{0}^{\frac{π}{2}} x^{2} \cos^{2} x d x . Then, I = \int_{0}^{\frac{π}{2}} x^{2} (\frac{1 + \cos 2 x}{2}) d x \Rightarrow I = \int_{0}^{\frac{π}{2}} (\frac{x^{2}}{2} + \frac{x^{2} \cos 2 x}{2}) d x \Rightarrow I = {[\frac{x^{3}}{6}]}_{0}^{\frac{π}{2}} + {[\frac{x^{2} \sin 2 x}{4}]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \frac{x}{2} \sin 2 x d x \Rightarrow I = {[\frac{x^{3}}{6}]}_{0}^{\frac{π}{2}} + {[\frac{x^{2} \sin 2 x}{4}]}_{0}^{\frac{π}{2}} - {[\frac{- x \cos 2 x}{4}]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} - 1 \frac{\cos 2 x}{2} d x \Rightarrow I = {[\frac{x^{3}}{6}]}_{0}^{\frac{π}{2}} + {[\frac{x^{2} \sin 2 x}{4}]}_{0}^{\frac{π}{2}} + {[\frac{x \cos 2 x}{4}]}_{0}^{\frac{π}{2}} - {[\frac{\sin 2 x}{4}]}_{0}^{\frac{π}{2}} \Rightarrow I = \frac{π^{3}}{48} - \frac{π}{8}$

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Question 32:

$\int_{1}^{2} \log x d x$

Answer:

$Let I = \int_{1}^{2} \log x d x . Then, I = \int_{1}^{2} 1 \log x d x Integrating by parts \Rightarrow I = {[x \log x]}_{1}^{2} - \int_{1}^{2} \frac{1}{x} x d x \Rightarrow I = {[x \log x]}_{1}^{2} - \int_{1}^{2} d x \Rightarrow I = {[x \log x]}_{1}^{2} - {[x]}_{1}^{2} \Rightarrow I = 2 \log 2 - 2 + 1 \Rightarrow I = 2 \log 2 - 1$

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Question 33:

$\int_{1}^{3} \frac{\log x}{{(x + 1)}^{2}} d x$

Answer:

$Let I = \int_{1}^{3} \frac{\log x}{{(1 + x)}^{2}} d x . Then, I = {[\frac{- 1}{1 + x} \log x]}_{1}^{3} - \int_{1}^{3} \frac{1}{x} (\frac{- 1}{x + 1}) d x \Rightarrow I = {[\frac{- 1}{1 + x} \log x]}_{1}^{3} + \int_{1}^{3} \frac{1}{x (x + 1)} d x \Rightarrow I = {[\frac{- 1}{1 + x} \log x]}_{1}^{3} + \int_{1}^{3} (\frac{1}{x} - \frac{1}{x + 1}) d x \Rightarrow I = {[\frac{- 1}{1 + x} \log x]}_{1}^{3} + {[\log x - \log (x + 1)]}_{1}^{3} \Rightarrow I = \frac{- 1}{4} \log 3 + \log 3 - \log 4 + \log 2 \Rightarrow I = \frac{3}{4} \log 3 - \log 2$

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Question 34:

$\int_{1}^{e} \frac{e^{x}}{x} (1 + x \log x) d x$

Answer:

$Let I = \int_{1}^{e} \frac{e^{x}}{x} (1 + x \log x) d x . Then, I = \int_{1}^{e} (\frac{e^{x}}{x} + e^{x} \log x) d x \Rightarrow I = \int_{1}^{e} \frac{e^{x}}{x} d x + \int_{1}^{e} e^{x} \log x d x Integrating first term by parts \Rightarrow I = {[\log x e^{x}]}_{1}^{e} - \int_{1}^{e} e^{x} \log x d x + \int_{1}^{e} e^{x} \log x d x \Rightarrow I = (\log e) e^{e} - 0 \Rightarrow I = e^{e}$

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Question 35:

$\int_{1}^{e} \frac{\log x}{x} d x$

Answer:

$Let I = \int_{1}^{e} \frac{\log x}{x} d x Let \log x = u \Rightarrow \frac{1}{x} d x = d u ∴ I = \int u d u \Rightarrow I = [\frac{u^{2}}{2}] \Rightarrow I = {[\frac{(\log x)^{2}}{2}]}_{1}^{e} \Rightarrow I = \frac{1}{2} - 0 \Rightarrow I = \frac{1}{2}$

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Question 36:

$\int_{e}^{e^{2}} \{\frac{1}{\log x} - \frac{1}{{(\log x)}^{2}}\} d x$

Answer:

$Let I = \int_{e}^{e^{2}} \{\frac{1}{\log x} - \frac{1}{{(\log x)}^{2}}\} d x . Then, I = \int_{e}^{e^{2}} 1 \frac{1}{\log x} d x - \int_{e}^{e^{2}} \frac{1}{{(\log x)}^{2}} d x Integrating by parts \Rightarrow I = \{{[\frac{x}{\log x}]}_{e}^{e^{2}} - \int_{e}^{e^{2}} \frac{- 1}{x {(\log x)}^{2}} x d x\} - \int_{e}^{e^{2}} \frac{1}{{(\log x)}^{2}} d x \Rightarrow I = {[\frac{x}{\log x}]}_{e}^{e^{2}} + \int_{e}^{e^{2}} \frac{1}{{(\log x)}^{2}} d x - \int_{e}^{e^{2}} \frac{1}{{(\log x)}^{2}} d x \Rightarrow I = {[\frac{x}{\log x}]}_{e}^{e^{2}} + 0 \Rightarrow I = \frac{e^{2}}{\log e^{2}} - \frac{e}{\log e} \Rightarrow I = \frac{e^{2}}{2 \log e} - \frac{e}{\log e} \Rightarrow I = \frac{e^{2}}{2} - e$

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Question 37:

$\int_{1}^{2} \frac{x + 3}{x (x + 2)} d x$

Answer:

$Let I = \int_{1}^{2} \frac{x + 3}{x (x + 2)} d x . Then, I = \int_{1}^{2} (\frac{x}{x (x + 2)} + \frac{3}{x (x + 2)}) d x \Rightarrow I = \int_{1}^{2} \frac{d x}{(x + 2)} + \int_{1}^{2} \frac{3}{x (x + 2)} d x \Rightarrow I = {[\log (x + 2)]}_{1}^{2} + \frac{3}{2} \int_{1}^{2} (\frac{1}{x} - \frac{1}{x + 2}) d x \Rightarrow I = {[\log (x + 2)]}_{1}^{2} + \frac{3}{2} {[\log x - \log (x + 2)]}_{1}^{2} \Rightarrow I = \log 4 - \log 3 + \frac{3}{2} [\log 2 - \log 4 - 0 + \log 3] \Rightarrow I = \log 4 - \log 3 + \frac{3}{2} [- \log 2 + \log 3] \Rightarrow I = 2 \log 2 - \log 3 + \frac{3}{2} \log 3 - \frac{3}{2} \log 2 \Rightarrow I = \frac{1}{2} \log 2 + \frac{1}{2} \log 3 \Rightarrow I = \frac{1}{2} (\log 2 + \log 3) \Rightarrow I = \frac{1}{2} \log 6$

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Question 38:

$\int_{0}^{1} \frac{2 x + 3}{5 x^{2} + 1} d x$

Answer:

$Let I = \int_{0}^{1} \frac{2 x + 3}{5 x^{2} + 1} d x . Then, I = \int_{0}^{1} \frac{2 x}{5 x^{2} + 1} d x + \int_{0}^{1} \frac{3}{5 x^{2} + 1} d x \Rightarrow I = \frac{1}{5} \int_{0}^{1} \frac{10 x}{5 x^{2} + 1} d x + 3 \int_{0}^{1} \frac{1}{{(\sqrt{5} x)}^{2} + 1^{2}} d x \Rightarrow I = \frac{1}{5} {[\log (5 x^{2} + 1)]}_{0}^{1} + \frac{3}{\sqrt{5}} {[\tan^{- 1} (\sqrt{5} x)]}_{0}^{1} \Rightarrow I = \frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{- 1} \sqrt{5}$

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Question 39:

$\int_{0}^{2} \frac{1}{4 + x - x^{2}} d x$

Answer:

$Let I = \int_{0}^{2} \frac{1}{4 + x - x^{2}} d x . Then, I = - \int_{0}^{2} \frac{1}{x^{2} - x - 4} d x \Rightarrow I = - \int_{0}^{2} \frac{1}{(x^{2} - x + \frac{1}{4}) - \frac{1}{4} - 4} d x = - \int_{0}^{2} \frac{1}{{(x - \frac{1}{2})}^{2} - \frac{17}{4}} d x = - \int_{0}^{2} \frac{1}{{(x - \frac{1}{2})}^{2} - {(\frac{\sqrt{17}}{2})}^{2}} d x = \int_{0}^{2} \frac{1}{- {(\frac{2 x - 1}{2})}^{2} + {(\frac{\sqrt{17}}{2})}^{2}} d x = \frac{1}{\sqrt{17}} {[\log (\frac{\sqrt{17} + 2 x - 1}{\sqrt{17} - 2 x + 1})]}_{0}^{2} = \frac{1}{\sqrt{17}} \{\log \frac{\sqrt{17} + 3}{\sqrt{17} - 3} - \log \frac{\sqrt{17} - 1}{\sqrt{17} + 1}\} = \frac{1}{\sqrt{17}} \{\log \frac{26 + 6 \sqrt{17}}{8} - \log \frac{18 - 2 \sqrt{17}}{16}\} = \frac{1}{\sqrt{17}} \{\log \frac{52 + 12 \sqrt{17}}{18 - 2 \sqrt{17}}\} = \frac{1}{\sqrt{17}} \{\log \frac{52 + 12 \sqrt{17}}{18 - 2 \sqrt{17}} \times \frac{18 + 2 \sqrt{17}}{18 + 2 \sqrt{17}}\} \Rightarrow I = \frac{1}{\sqrt{17}} \log \frac{1344 + 320 \sqrt{17}}{256} \Rightarrow I = \frac{1}{\sqrt{17}} \log \frac{21 + 5 \sqrt{17}}{4}$

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Question 40:

$\int_{0}^{1} \frac{1}{2 x^{2} + x + 1} d x$

Answer:

$Let I = \int_{0}^{1} \frac{1}{2 x^{2} + x + 1} d x . Then, I = \frac{1}{2} \int_{0}^{1} \frac{1}{x^{2} + \frac{x}{2} + \frac{1}{2}} d x I = \frac{1}{2} \int_{0}^{1} \frac{1}{(x^{2} + \frac{x}{2} + \frac{1}{16}) - \frac{1}{16} + \frac{1}{2}} d x \Rightarrow I = \frac{1}{2} \int_{0}^{1} \frac{1}{{(x + \frac{1}{4})}^{2} + \frac{7}{16}} d x \Rightarrow I = \frac{1}{2} \times \frac{4}{\sqrt{7}} {[\tan^{- 1} (\frac{x + \frac{1}{4}}{\frac{\sqrt{7}}{4}})]}_{0}^{1} \Rightarrow I = \frac{2}{\sqrt{7}} (\tan^{- 1} \frac{5}{\sqrt{7}} - \tan^{- 1} \frac{1}{\sqrt{7}})$

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Question 41:

$\int_{0}^{1} \sqrt{x (1 - x)} d x$

Answer:

$Let I = \int_{0}^{1} \sqrt{x (1 - x)} d x . Then, I = \int_{0}^{1} \sqrt{\frac{1}{4} - {(x - \frac{1}{2})}^{2}} d x \Rightarrow I = \frac{1}{2} \int_{0}^{1} \sqrt{1 - \frac{{(x - \frac{1}{2})}^{2}}{\frac{1}{4}}} d x \Rightarrow I = \frac{1}{2} \int_{0}^{1} \sqrt{1 - {(\frac{x - \frac{1}{2}}{\frac{1}{2}})}^{2}} d x Let (\frac{x - \frac{1}{2}}{\frac{1}{2}}) = \sin u \Rightarrow 2 d x = \cos u d u ∴ I = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{1 - \sin^{2} u} \cos u d u \Rightarrow I = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} u d u \Rightarrow I = \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{\cos 2 u + 1}{2}) d u \Rightarrow I = \frac{1}{8} {[\frac{\sin 2 u}{2} + u]}_{- \frac{π}{2}}^{\frac{π}{2}} \Rightarrow I = \frac{1}{8} [\frac{π}{2} + \frac{π}{2}] \Rightarrow I = \frac{π}{8}$

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Question 42:

$\int_{0}^{2} \frac{1}{\sqrt{3 + 2 x - x^{2}}} d x$

Answer:

$Let I = \int_{0}^{2} \frac{1}{\sqrt{3 + 2 x - x^{2}}} d x . Then, I = \int_{0}^{2} \frac{1}{\sqrt{- x^{2} + 2 x - 1 + 1 + 3}} d x \Rightarrow I = \int_{0}^{2} \frac{1}{\sqrt{- {(x - 1)}^{2} + 4}} d x \Rightarrow I = {[\sin^{- 1} \frac{(x - 1)}{2}]}_{0}^{2} \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} (- \frac{1}{2}) \Rightarrow I = 2 \sin^{- 1} \frac{1}{2} \Rightarrow I = 2 \times \frac{π}{6} = \frac{π}{3}$

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Question 43:

$\int_{0}^{4} \frac{1}{\sqrt{4 x - x^{2}}} d x$

Answer:

$Let I = \int_{0}^{4} \frac{1}{\sqrt{4 x - x^{2}}} d x . Then, I = \int_{0}^{4} \frac{1}{\sqrt{4 x - x^{2} - 4 + 4}} d x \Rightarrow I = \int_{0}^{4} \frac{1}{\sqrt{- {(x - 2)}^{2} + 4}} d x \Rightarrow I = {[\sin^{- 1} \frac{(x - 2)}{2}]}_{0}^{4} \Rightarrow I = (\sin^{- 1} 1 - \sin^{- 1} (- 1)) \Rightarrow I = 2 \sin^{- 1} 1 \Rightarrow I = 2 \frac{π}{2} = π$

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Question 44:

$\int_{- 1}^{1} \frac{1}{x^{2} + 2 x + 5} d x$

Answer:

$Let I = \int_{- 1}^{1} \frac{1}{x^{2} + 2 x + 5} d x . Then, I = \int_{- 1}^{1} \frac{1}{(x^{2} + 2 x + 1) + 4} d x \Rightarrow I = \int_{- 1}^{1} \frac{1}{{(x + 1)}^{2} + 2^{2}} d x \Rightarrow I = \frac{1}{2} {[\tan^{- 1} \frac{(x + 1)}{2}]}_{- 1}^{1} \Rightarrow I = \frac{1}{2} (\tan^{- 1} 1 - \tan^{- 1} 0) \Rightarrow I = \frac{1}{2} (\frac{π}{4}) \Rightarrow I = \frac{π}{8}$

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Question 45:

$\int_{1}^{4} \frac{x^{2} + x}{\sqrt{2 x + 1}} d x$

Answer:

$Let I = \int_{1}^{4} \frac{x^{2} + x}{\sqrt{2 x + 1}} d x . Let 2 x + 1 = u \Rightarrow x = \frac{u - 1}{2} \Rightarrow d x = \frac{d u}{2} ∴ I = \int \frac{{(\frac{u - 1}{2})}^{2} + \frac{u - 1}{2}}{\sqrt{u}} \frac{d u}{2} \Rightarrow I = \frac{1}{8} \int \frac{u^{2} + 1 - 2 u + 2 u - 2}{\sqrt{u}} d u = \frac{1}{8} \int \frac{(u^{2} - 1)}{\sqrt{u}} d u = \frac{1}{8} \int (u^{\frac{3}{2}} - u^{- \frac{1}{2}}) d u = \frac{1}{8} [\frac{2 u^{\frac{5}{2}}}{5} - \frac{2 u^{\frac{1}{2}}}{1}] = \frac{1}{8} {[\frac{2 {(2 x + 1)}^{\frac{5}{2}}}{5} - \frac{2 {(2 x + 1)}^{\frac{1}{2}}}{1}]}_{1}^{4} = \frac{1}{8} [\frac{2}{5} \times 243 - 6 - \frac{2}{5} \times 9 \sqrt{3} + 2 \sqrt{3}] \Rightarrow I = \frac{1}{8} [\frac{456}{5} - \frac{8 \sqrt{3}}{5}] \Rightarrow I = \frac{57 - \sqrt{3}}{5}$

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Question 46:

$\int_{0}^{1} x {(1 - x)}^{5} d x$

Answer:

$Let I = \int_{0}^{1} x {(1 - x)}^{5} d x . Then, I = \int_{0}^{1} (x - 1 + 1) {(1 - x)}^{5} d x \Rightarrow I = \int_{0}^{1} [- {(1 - x)}^{6} + {(1 - x)}^{5}] d x \Rightarrow I = {[\frac{{(1 - x)}^{7}}{7}]}_{0}^{1} - {[\frac{{(1 - x)}^{6}}{6}]}_{0}^{1} \Rightarrow I = - \frac{1}{7} + \frac{1}{6} \Rightarrow I = \frac{1}{42}$

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Question 47:

$\int_{1}^{2} (\frac{x - 1}{x^{2}}) e^{x} d x$

Answer:

$Let I = \int_{1}^{2} (\frac{x - 1}{x^{2}}) e^{x} d x . Then, I = \int_{1}^{2} (\frac{e^{x}}{x} - \frac{e^{x}}{x^{2}}) d x \Rightarrow I = \int_{1}^{2} \frac{e^{x}}{x} d x - \int_{1}^{2} \frac{e^{x}}{x^{2}} d x Integrating first term by parts I = \{{[\frac{e^{x}}{x}]}_{1}^{2} - \int_{1}^{2} \frac{- 1}{x^{2}} e^{x} d x\} - \int_{1}^{2} \frac{e^{x}}{x^{2}} d x \Rightarrow I = {[\frac{e^{x}}{x}]}_{1}^{2} + \int_{1}^{2} \frac{e^{x}}{x^{2}} d x - \int_{1}^{2} \frac{e^{x}}{x^{2}} d x \Rightarrow I = {[\frac{e^{x}}{x}]}_{1}^{2} \Rightarrow I = \frac{e^{2}}{2} - e$

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Question 48:

$\int_{0}^{1} (x e^{2 x} + \sin \frac{πx}{2}) d x$

Answer:

$Let I = \int_{0}^{1} (x e^{2 x} + \sin \frac{πx}{2}) d x . Then, I = \int_{0}^{1} x e^{2 x} d x + \int_{0}^{1} \sin \frac{π x}{2} d x Integrating first term by parts I = {[x \frac{e^{2 x}}{2}]}_{0}^{1} - \int_{0}^{1} 1 \frac{e^{2 x}}{2} d x + {[- \frac{\cos \frac{πx}{2}}{\frac{π}{2}}]}_{0}^{1} \Rightarrow I = {[x \frac{e^{2 x}}{2}]}_{0}^{1} - {[\frac{e^{2 x}}{4}]}_{0}^{1} - \frac{2}{π} {[\cos \frac{πx}{2}]}_{0}^{1} \Rightarrow I = \frac{e^{2}}{2} - \frac{e^{2}}{4} + \frac{1}{4} + \frac{2}{π} \Rightarrow I = \frac{e^{2}}{4} + \frac{1}{4} + \frac{2}{π}$

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Question 49:

$\int_{0}^{1} (x e^{x} + \cos \frac{πx}{4}) d x$

Answer:

$Let I = \int_{0}^{1} (x e^{x} + \cos \frac{π x}{4}) d x . Then, I = \int_{0}^{1} x e^{x} d x + \int_{0}^{1} \cos \frac{π x}{4} d x Integrating first term by parts I = \{{[x e^{x}]}_{0}^{1} - \int_{0}^{1} 1 e^{x} d x\} + {[\frac{\sin \frac{π x}{4}}{\frac{π}{4}}]}_{0}^{1} \Rightarrow I = {[x e^{x}]}_{0}^{1} - {[e^{x}]}_{0}^{1} + {[\frac{\sin \frac{π x}{4}}{\frac{π}{4}}]}_{0}^{1} \Rightarrow I = e - e + 1 + \frac{4}{π} \sin \frac{π}{4} \Rightarrow I = 1 + \frac{4}{π \sqrt{2}} \Rightarrow I = 1 + \frac{2 \sqrt{2}}{π}$

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.

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Question 50:

$\int_{π / 2}^{π} e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x$

Answer:

$Let I = \int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - \sin x}{1 - \cos x}) d x . Then, I = \int_{\frac{π}{2}}^{π} e^{x} (\frac{1 - 2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin^{2} \frac{x}{2}}) d x [As, \sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}, \cos A = 1 - 2 \sin^{2} \frac{A}{2}] \Rightarrow I = \int_{\frac{π}{2}}^{π} e^{x} (\frac{1}{2} {cosec}^{2} \frac{x}{2} - \cot \frac{x}{2}) d x \Rightarrow I = \int_{\frac{π}{2}}^{π} \frac{1}{2} e^{x} {cosec}^{2} \frac{x}{2} d x - \int_{\frac{π}{2}}^{π} e^{x} \cot \frac{x}{2} d x Integrating second term by parts I = \{- {[e^{x} \cot \frac{x}{2}]}_{\frac{π}{2}}^{π} - \int_{\frac{π}{2}}^{π} \frac{1}{2} e^{x} {cosec}^{2} \frac{x}{2} d x\} + \int_{\frac{π}{2}}^{π} \frac{1}{2} e^{x} {cosec}^{2} \frac{x}{2} d x \Rightarrow I = - [0 - e^{\frac{π}{2}}] \Rightarrow I = e^{\frac{π}{2}}$