integrate 1/sinx-sin2x w r t x - Maths - Integrals

Question

integrate 1/sinx-sin2x w r t x ??

ramandeep.singh · Accepted Answer

$I = \int \frac{1}{\sin x - \sin 2 x} ⅆ x = \int \frac{1}{\sin x (1 - 2 \cos x)} ⅆ x$

Let

$t = \cos x ∴ d t = - \sin x d x \Rightarrow d x = \frac{- d t}{\sin x} \sin x = \sqrt{1 - \cos^{2} x} = \sqrt{1 - t^{2}}$

$∴ I = \int \frac{- 1}{(1 - t^{2}) (1 - 2 t)} ⅆ t \frac{- 1}{(1 - t^{2}) (1 - 2 t)} = \frac{A}{1 + t} + \frac{B}{1 - t} + \frac{C}{1 - 2 t}$ (Using partial fractions)

On solving, we get

$A = \frac{- 1}{6}, B = \frac{3}{6}, C = \frac{- 8}{6}$

$∴ I = \frac{- 1}{6} \int \frac{1}{1 + t} ⅆ t + \frac{3}{6} \int \frac{1}{1 - t} ⅆ t - \frac{8}{6} \int \frac{1}{1 - 2 t} ⅆ t = \frac{- 1}{6} \log | 1 + t | - \frac{3}{6} \log | 1 - t | - \frac{8}{6 (- 2)} \log | 1 - 2 t | + C = \frac{- 1}{6} \log | 1 + \cos x | - \frac{1}{2} \log | 1 - \cos x | + \frac{2}{3} \log | 1 - 2 \cos x | + C$

integrate 1/sinx-sin2x w.r.t x ??