$\displaystyle

\int sin{^2}(x)cos{^4}(x)

$

i did it all the way to

$\displaystyle

\frac {1}{8} \int sin{^2}(2x)+sin{^2}(2x)cos(2x)

$

then i split it up and have no idea what happens next...

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- Feb 26th 2008, 12:46 PMsimsima_1integration helpppppp
$\displaystyle

\int sin{^2}(x)cos{^4}(x)

$

i did it all the way to

$\displaystyle

\frac {1}{8} \int sin{^2}(2x)+sin{^2}(2x)cos(2x)

$

then i split it up and have no idea what happens next... - Feb 26th 2008, 12:49 PMJhevon
hopefully your last line is right, i didn't check. it looks a little fishy to me.

anyway, note that $\displaystyle \sin^2 2x = \frac {1 - \cos 4x}2$. so you can easily integrate the first part

for the second part, use the substitution $\displaystyle u = \sin 2x$