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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

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