integral of ( e2x( cosx + 7sinx )dx) = e2x(g(x)) + c where c ia s constant of - Maths - Inverse Trigonometric Functions

Question

integral of ( e2x( cosx + 7sinx )dx) =
e2x(g(x)) + c where c ia s constant of
integration, then g(0) + g(pi/2) = :
a) 0 b) 1 c) 2 d) -1 No links

Krishna Murari · Accepted Answer

integral of ( e2x( cosx + 7sinx )dx) = e2x(g(x)) + c where c ia s constant of integration, then g(0) + g(pi/2) = : a) 0 b) 1 c) 2 d) -1 . No links