# Darn integral

1. Nov 16, 2009

### Void123

1. The problem statement, all variables and given/known data

I just can't crack the integral of (sin(x))^6 for some reason.

What is the exact solution to this? This is not really a homework question, as an immediate reference to an integral table would be sufficient. But I just need it right away. Thanks.

2. Relevant equations

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3. The attempt at a solution

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2. Nov 16, 2009

### jgens

Use integration by parts repeatedly.

3. Nov 16, 2009

### HallsofIvy

Staff Emeritus
Or use trig identities. $cos(2x)= cos^2(x)- sin^2(x)= 1- 2sin^2(x)$ so $sin^2(x)= (1/2)(1- cos(2x))$. Then $sin^6(x)= (sin^2(x))^3= (1/2)^3(1- cos(2x))^3$$= (1/8)(1- 3cos(2x)$$+ 3cos^2(2x)+ cos^3(2x))$.

The integral of 1- 3cos(2x) is straightforward. The integral of $cos^3(2x)$ can be done by writing it as $cos^3(2x)= cos(2x)(1- sin^2(2x))$ and using the substitution u= sin(2x). The integral of $cos^2(2x)$ can be done by using the trig identity $cos^2(2x)= (1/2)(1+ cos(4x))$.