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Can somebody help me calculate me the following integral:

$\int_{0}^1 x(\arcsin x)^2dx$ Please help

Thank you in advance

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up vote 0 down vote accepted

Make following substitution: $u=\arcsin x.$ Then $x=\sin{u}$ and \begin{gather} \int\limits_{0}^1 x(\arcsin x)^2dx=\int\limits_{0}^{\tfrac{\pi}{2}}{u^2\sin{u}\,d(\sin{u})}=\int\limits_{0}^{\tfrac{\pi}{2}}{u^2\sin{u}\cos{u}\,d{u}}=\\ \frac{1}{2}\int\limits_{0}^{\tfrac{\pi}{2}}{u^2\sin{2u}\,d{u}}=\frac{1}{16}\int\limits_{0}^{\tfrac{\pi}{2}}{(2u)^2\sin{2u}\,d{(2u)}}=\frac{1}{16}\int\limits_{0}^{\pi}{v^2 \sin{v}\,dv}. \end{gather} Integrating last integral by parts (twice), as suggest Shafat Arbaz Alam, gives the desired result.

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Have you tried integration by parts?

$$\int u \,dv = uv - \int v \,du$$

You may need more than one iteration.

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