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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.

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2 Answers

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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?

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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}

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