Integration of trigonometric Identities.

**davewhite123** · September 7th 2009, 12:10 AM

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

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

**red_dog** · September 7th 2009, 01:10 AM

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

Then

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

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

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

The first integral is easy.

In the second let $u-\sin x$

In the third let $u=\cos x$

In the fourth let $u=\sin x$

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