# Thread: Integration of trigonometric Identities.

1. ## Integration of trigonometric Identities.

Problem:
$\displaystyle \int sinx(1-sin2x)^{3}dx$

I have been trying this calculus problem for so long and don't seem to solve it, so please help me solve it?

Thanks for any help

2. $\displaystyle (1-\sin 2x)^3=(1-2\sin x\cos x)^3=1-6\sin x\cos x+12\sin^2x\cos^2x-8\sin^3x\cos^3x$

Then

$\displaystyle \int\sin x(1-\sin 2x)^3dx=\int(\sin x-6\sin^2x\cos x+12\sin^3x\cos^2x-8\sin^4x\cos^3x)dx=$

$\displaystyle =\int\sin xdx-6\int\sin^2x\cos xdx+$

$\displaystyle +12\int(1-\cos^2x)\cos^2x\sin xdx-8\int\sin^4x(1-\sin^2x)\cos xdx$

The first integral is easy.

In the second let $\displaystyle u-\sin x$

In the third let $\displaystyle u=\cos x$

In the fourth let $\displaystyle u=\sin x$