$\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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Originally Posted by simsima_1 $\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... 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 $\displaystyle u = \sin 2x$
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