Question

Hi I need help integrating (arcsin(x))^2 dx
I was told to integrate by parts 2 or 3 times

Answer 1

\[\int \arcsin^2x\,dx\]First thing I would do is set \(u=\arcsin x\) so that \(x=\sin u\) (notice that this implicitly places some restrictions on the domain of \(u\)), then \(dx=\cos u\,du\) and you have
\[\int \arcsin^2x\,dx=\int u^2\cos u\,du \]I suspect by now you've had practice dealing with integrals like this? Integrating by parts twice will reduce the power of \(x\).

Alternatively, you can implement integration by parts immediately with the following setup:
\[\begin{matrix}f=\arcsin^2x&&&dg=dx\\
df=\dfrac{2\arcsin x}{\sqrt{1-x^2}}\,dx&&&g=x\end{matrix}\]which gives
\[\int \arcsin^2x\,dx=x\arcsin^2x-2\int\frac{x\arcsin x}{\sqrt{1-x^2}}\,dx\]where you can make the same substitution that I first used, i.e. \(u=\arcsin x\), so \(\sin u=x\) and \(du=\dfrac{dx}{\sqrt{1-x^2}}\), giving
\[\int\frac{x\arcsin x}{\sqrt{1-x^2}}\,dx=\int u\sin u\,du\]which takes one round of IBP, or you can treat the integral with IBP again. Let
\[\begin{matrix}f=\arcsin x&&&dg=\dfrac{x}{\sqrt{1-x^2}}\,dx\\
df=\dfrac{dx}{\sqrt{1-x^2}}&&&g=-\sqrt{1-x^2}\end{matrix}\]Then
\[\begin{align*}\int \arcsin^2x\,dx&=x\arcsin^2x-2\left(-\sqrt{1-x^2}\arcsin x+\int dx\right)\\[1ex]
&=x\arcsin^2x+2\sqrt{1-x^2}\arcsin x-2\int dx\end{align*}\]