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.
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
$$ \int x\arcsin x\mathrm dx=\frac12x^2\arcsin x-\frac14\arcsin x+\frac14x\sqrt{1-x^2}+C $$
\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?
