How do I integrate (sinx)^6 ?

**s3a** · February 27th 2010, 05:41 AM

I could integrate (sinx)^4 by doing (sin^2 x)^2 for example and then using the half angle formula (1 - cos(2x))/2 but I can't do that for (sinx)^6 and Wolfram tells me to use a reduction formula which is not something I am supposed to know yet.

So can someone please help me to do this integral? It's not the actual problem I am doing but is part of the problem that I think I am stuck on. (My real homework problem is: the Integral of (sinx)^4 * (cosx)^2 dx.

Any help would be greatly appreciated!
Thanks in advance!

**skeeter** · February 27th 2010, 06:16 AM

Originally Posted by s3a

I could integrate (sinx)^4 by doing (sin^2 x)^2 for example and then using the half angle formula (1 - cos(2x))/2 but I can't do that for (sinx)^6 and Wolfram tells me to use a reduction formula which is not something I am supposed to know yet.

So can someone please help me to do this integral? It's not the actual problem I am doing but is part of the problem that I think I am stuck on. (My real homework problem is: the Integral of (sinx)^4 * (cosx)^2 dx.

Any help would be greatly appreciated!
Thanks in advance!

repeated use of the double angle power reduction identity ...

$\int \sin^2{x} \cdot \sin^2{x} \cdot \cos^2{x} \, dx<br />$

$\int \left[\frac{1-\cos(2x)}{2}\right] \cdot \left[\frac{1-\cos(2x)}{2}\right] \cdot \left[\frac{1+\cos(2x)}{2}\right] \, dx$

$\int \left[\frac{1-\cos(2x)}{2}\right] \cdot \left[\frac{1-\cos^2(2x)}{4}\right] \, dx$

$\int \left[\frac{1-\cos(2x)}{2}\right] \cdot \left[\frac{\sin^2(2x)}{4}\right] \, dx$

$\frac{1}{8}\int \sin^2(2x) - \sin^2(2x) \cdot \cos(2x) \, dx$

$\frac{1}{8} \int \sin^2(2x) \, dx - \frac{1}{8}\int \sin^2(2x) \cdot \cos(2x) \, dx$

$\frac{1}{16} \int 1 - \cos(4x) \, dx - \frac{1}{8} \int \sin^2(2x) \cdot \cos(2x) \, dx$

$\frac{1}{16}\left[x - \frac{\sin(4x)}{4}\right] - \frac{1}{16} \cdot \frac{\sin^3(2x)}{3} + C$

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