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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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  • $\begingroup$ In your last integral, integrate by parts by differentiating x and integrating the other stuff $\endgroup$ Jan 24 '21 at 9:45
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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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    $\begingroup$ 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. $\endgroup$
    – Ski Mask
    Jan 24 '21 at 10:19
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    $\begingroup$ @SkiMask Exactly. Well done! $\endgroup$
    – Robert Z
    Jan 24 '21 at 10:22
  • $\begingroup$ Using $x=\sin t$ at the start would be easier. $\endgroup$
    – J.G.
    Jan 24 '21 at 16:34
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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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