Integrating $\int x \arcsin x \,dx$ by parts Solving the integral $$\int x \arcsin x\,dx$$
I know I have to solve integral using by parts
$$\int u\,dv = uv-  \int v\,du$$
$$\int x \arcsin x \,dx$$
Let $u = \arcsin x$ and $dv = x\,dx$
so $du = \dfrac1{\sqrt{1-x^2}} \, dx$
and $dv = x^2$.
but here is where I am stuck. because when I plug the variables into equation it becomes confusing on what to do next.
ps. New to using latex and still haven't perfected it so I apologize for formatting.
 A: Using OP's notation, we have $v=x^2/2$ and $\mathrm du=\mathrm dx/\sqrt{1-x^2}$, so that
$$
\int x\arcsin x\mathrm dx=\frac12x^2\arcsin x-\frac12\underbrace{\int{x^2\over\sqrt{1-x^2}}\mathrm dx}_I
$$
For the latter part, we have
$$
\begin{aligned}
I
&=\int{x^2\over\sqrt{1-x^2}}\mathrm dx \\
&=\int{x^2-1+1\over\sqrt{1-x^2}}\mathrm dx \\
&=-\int\sqrt{1-x^2}\mathrm dx+\int{\mathrm dx\over\sqrt{1-x^2}} \\
&=\arcsin x-\int\sqrt{1-x^2}\mathrm dx
\end{aligned}
$$
For the remaining integral, it follows from yet another IBP that
$$
\begin{aligned}
\int\sqrt{1-x^2}\mathrm dx
&=x\sqrt{1-x^2}+\int{x^2\over\sqrt{1-x^2}}\mathrm dx \\
&=x\sqrt{1-x^2}+I
\end{aligned}
$$
Plugging this back in, we have
$$
2I=\arcsin x-x\sqrt{1-x^2}+C
$$
Putting everything together, we get
$$
\int x\arcsin x\mathrm dx=\frac12x^2\arcsin x-\frac14\arcsin x+\frac14x\sqrt{1-x^2}+C
$$
A: \begin{gather*}
I=\int x\arcsin x\,dx\\
\text{Let } x=\sin \theta \\
dx=\cos \theta \,d\theta \\
I=\int \sin \theta \cos \theta \,d\theta \cdot \theta =\frac{1}{2}\int \theta \sin 2\theta \,d\theta \\
\text{By integration by parts method,}\\
\int uv\,dx=u\int v\,dx-\int \frac{du}{dx}\left(\int v\,dx\right) dx\\
\text{Take } u=\theta \text{ and } v=\sin 2\theta .
\end{gather*}
Can you go ahead and take it from here?
