# Evaluate $\int x \arcsin(x)dx$.

Evaluate: $$\int x \arcsin(x)dx$$

My first guess was $$u$$ substitution but that didn't get me very far. I think using integration by parts is the correct way. Here's my attempt:

$$u = \arcsin(x), v' = x \Longrightarrow \int x \arcsin(x)dx = \arcsin(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2\sqrt{1-x^2}}dx$$

But I am stuck on how to continue from here.

• In your last integral, integrate by parts by differentiating x and integrating the other stuff Jan 24 '21 at 9:45

You are on the right track. As regards the remaining integral, note that $$\int \frac{x^2}{2\sqrt{1-x^2}}\,dx=\frac{1}{2}\int \frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx =\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}}\,dx-\frac{1}{2}\int \sqrt{1-x^2}\,dx.$$ The first integral is easy, whereas for the second one you may use the substitution $$x=\sin(t)$$.

Can you take it from here?

• The first integral would then become $\frac{1}{2} \arcsin(x)$ and for the second integral you would have to use $u-$substitution and the solution to it would be $\frac{\arcsin(x)+x\sqrt{1-x^2}}{4}$. Combining the solutions would then give me the solution for the original integral. Jan 24 '21 at 10:19
• @SkiMask Exactly. Well done! Jan 24 '21 at 10:22
• Using $x=\sin t$ at the start would be easier.
– J.G.
Jan 24 '21 at 16:34

Integrate by parts twice

\begin{align} \int x \arcsin x\> dx &=\frac12 x^2 \arcsin x -\frac12 \int \frac{x^2}{\sqrt{1-x^2}}\,dx\\ &=\frac12 x^2 \arcsin x+\frac14 \int \frac x{\sqrt{1-x^2}} \>d\left( 1-x^2\right)\\ &=\frac12 x^2 \arcsin x +\frac14 x \sqrt{1-x^2} - \frac14\arcsin x+ C\\ \end{align}