Integration of trig functions

**nyasha** · March 14th 2009, 10:11 PM

$\int sin^2xcos^4xdx$

Attempt to solution:

$sin^2x=1-cos^2x$

$\int(1-cos^2x)(cos^4x)dx$

$\int(cos^4x-cos^6x)dx$ ( I am stuck here, how l do evaluate this integral)

**Jhevon** · March 14th 2009, 10:18 PM

Originally Posted by nyasha

$\int sin^2xcos^4xdx$

recall that $\sin^2 x = \frac {1 - \cos 2x}2$ and $\cos^2 x = \frac {1 + \cos 2x}2$

**nyasha** · March 14th 2009, 10:45 PM

Originally Posted by Jhevon

recall that $\sin^2 x = \frac {1 - \cos 2x}2$ and $\cos^2 x = \frac {1 + \cos 2x}2$

$\frac{1}{2}\int(1-cos^2x)(cos^4x)dx<br />$

$\frac{1}{2}\int(cos^4x-cos^6x)dx$

$\frac{1}{2}\int cos^4xdx-\frac{1}{2}\int cos^6xdx$

$\frac{1}{8}\int (1+cos2x)^2-\frac{1}{16}\int(1+cos2x)^3$

I'm l now on the right path

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