Triple-angle formulae in integration

**gamboo** · May 7th 2009, 09:07 AM

∫sin^2(3x) dx

**red_dog** · May 7th 2009, 09:41 AM

$\int\sin^23xdx=\int\frac{1-\cos 6x}{2}dx=\frac{1}{2}\int dx-\frac{1}{2}\int\cos 6xdx=\frac{x}{2}-\frac{\sin 6x}{12}+C$

**gamboo** · May 7th 2009, 02:37 PM

I dont understand:

$\int\sin^23xdx=\int\frac{1-\cos 6x}{2}dx$

more detail please

**skeeter** · May 7th 2009, 02:43 PM

Originally Posted by gamboo

I dont understand:

$\int\sin^23xdx=\int\frac{1-\cos 6x}{2}dx$

more detail please

power reduction identity ...

$\sin^2{u} = \frac{1 - \cos(2u)}{2}$

derived from the double angle identity for cosine ...

$\cos(2u) = 1 - 2\sin^2{u}<br />$

**gamboo** · May 7th 2009, 03:33 PM

Thanks to both of you.

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