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