Tricky integral

**mmattson07** · September 23rd 2009, 03:25 PM

$\int_0^\pi \! 29sin^4(3t) \, dt.$

Not sure if I have to substitute just once or twice. Any suggestions?

**skeeter** · September 23rd 2009, 03:36 PM

Originally Posted by mmattson07

$\int_0^\pi \! 29sin^4(3t) \, dt.$

Not sure if I have to substitute just once or twice. Any suggestions?

$\sin^4{u} =$

$(\sin^2{u})^2 =$

$\left[\frac{1 - \cos(2u)}{2}\right]^2 =$

$\frac{1}{4}\left[1 - 2\cos(2u) + \cos^2(2u)\right] =$

$\frac{1}{4}\left[1 - 2\cos(2u) + \frac{1 + \cos(4u)}{2}\right] =$

$\frac{3}{8} - \frac{\cos(2u)}{2} + \frac{\cos(4u)}{8}$

**mmattson07** · September 23rd 2009, 03:41 PM

I see how you got this and i know $sin^2x=1-cos^2x$ but how did you get $1-cos(2u)/2$

**Stev381** · September 23rd 2009, 03:43 PM

Half-Angle formula.

**mr fantastic** · September 23rd 2009, 03:47 PM

Originally Posted by mmattson07

I see how you got this and i know $sin^2x=1-cos^2x$ but how did you get $1-cos(2u)/2$

Before attempting integrals like the one you posted you should ensure that you are thoroughly familiar with all required trigonometric identities. Otherwise you have no chance.

For example, you should know that $\cos (2u) = \cos^2 u - \sin^2 u = 1 - 2 \sin^2 u$ ....

**mmattson07** · September 23rd 2009, 03:56 PM

Yeah maybe that would help. He skipped a few algebraic steps so i was lost for a second.

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