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I don't know how to integrate $\int x^2\sin^2(x)\,\mathrm dx$. I know that I should do it by parts, where I have $ u=x^2\quad v'=\sin^2x \\ u'=2x \quad v={-\sin x\cos x+x\over 2}$ and now I have

$ \int x^2\sin^2(x)\,\mathrm dx = {-\sin x\cos x+x\over 2} x^2 - \int 2x{-\sin x\cos x+x\over 2}\,\mathrm dx\\ ={-\sin x\cos x+x\over 2} x^2 - \int x(-\sin x\cos x+x)\,\mathrm dx\\ $ so I have to calculate $ \int x(-\sin x\cos x+x)\,\mathrm dx=-\int x\sin x\cos x\,\mathrm dx+\int x^2\,\mathrm dx$

I know that $\int x^2\,\mathrm dx = {1 \over 3}x^3+C$ but I don't know what should I do with $\int x\sin x\cos x \,\mathrm dx$ Should i use parts again or what? Please help.

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    Yep, you should integrate by parts again. You might be interested in the trig identity $\sin(2x)=2\sin x\cos x$.2012-11-17

2 Answers 2

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We have $\sin x \cos x=\frac{1}{2}\sin 2x$. Now another integration by parts will do it.

It may be marginally easier to note at the beginning that $\cos 2x=1-2\sin^2 x$. So we want to integrate $\dfrac{1}{2}x^2(1-\cos 2x)$, that is, $\dfrac{x^2}{2}+\dfrac{x^2}{2}\cos 2x$, which looks more familiar.

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    Well of course! Thank you for your help, greetings from Poland :)2012-11-17
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All you need is in @Andre's answer but, I'd like to note you a simple way for solving the latter integral:

enter image description here

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    @amWhy: Thanks Amy. I made it wit office Word2013-04-05