# integrate sin^4(x)

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• May 14th 2009, 11:58 PM
quickzxxx
integrate sin^4(x)
hey guys im trying to integrate sin^4(x) can someone check over this and tell me if its correct:

= sin(4x)/32 - sin(2x)/4 + 3x/8

is that the correct answer for the integration of sin^4(x)
• May 15th 2009, 12:15 AM
Moo
Hello,
Quote:

Originally Posted by quickzxxx
hey guys im trying to integrate sin^4(x) can someone check over this and tell me if its correct:

= sin(4x)/32 - sin(2x)/4 + 3x/8

is that the correct answer for the integration of sin^4(x)

your answer is correct (don't forget the integration constant !)
• May 15th 2009, 12:25 AM
Spec
You don't have to use de Moivre's formula to solve this.

$\sin^4 x = \frac{1-\cos 2x}{2}\cdot \frac{1-\cos 2x}{2}=\frac{1}{4}(1-2\cos 2x + \cos^2 2x)=$ $\frac{1}{4}\left(1-2\cos 2x + \frac{1}{2}(1+\cos 4x)\right)=\frac{1}{8}\cos 4x - \frac{1}{2} \cos 2x + \frac{3}{8}$