Does anyone know how to get the integral of (sinx)^5 * cosx ?

Question

Does anyone know how to get the integral of (sinx)^5 * cosx ?

Answer 1

let u = sin(x)

du = cos(x) dx

∫ (sin(x))^5 * cos(x) dx

= ∫ u^5 du

= u^6/6 + C

= (sin(x))^6 / 6 + C

Answer 2

I = ∫ (sin x)^5 cos x dx

Let u = sin x

du = cos x dx

I = ∫ u^5 du

I = u^6/6 + C

I = (sin x)^6 / 6  + C

Answer 3

use u-substitution:

∫ (sin x)^5 * cos(x) dx

u = sin x

du = cos x dx

∫ u^5 du

= (1/6) u^6 + C

= (1/6) (sin x)^6 + C, where C is the constant due to the indefinite integration