Can somebody help me calculate me the following integral:
$\int_{0}^1 x(\arcsin x)^2dx$ Please help
Thank you in advance
Can somebody help me calculate me the following integral:
$\int_{0}^1 x(\arcsin x)^2dx$ Please help
Thank you in advance
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.
Have you tried integration by parts?
$\int u \,dv = uv - \int v \,du$
You may need more than one iteration.