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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 $\sin(2x)=2\sin x\cos x$. –  icurays1 Nov 17 '12 at 17:47
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2 Answers

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 :) –  Emilia Nov 17 '12 at 17:50
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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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Nice table! Great visual...+1 –  amWhy Apr 5 '13 at 1:34
    
@amWhy: Thanks Amy. I made it wit office Word –  B. S. Apr 5 '13 at 13:29
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