# Evaluate $\int (\arcsin(x))^2 dx$

How to evaluate this?

$$\int \left(\arcsin(x)\right)^2 dx$$

Integral by parts gives me

$$\int\left(\arcsin(x)\right)^2 dx =-\int2x\arcsin(x)\frac{1}{\sqrt{1-x^2}}dx+x\arcsin(x)$$

Let $u = \arcsin(x)$. Then $\frac{du}{dx} = \frac{1}{\sqrt{1-x^2}}$. Thus, the integral on the right is

$$-2\int \frac{x}{\sqrt{1-x^2}}\arcsin(x)dx=-2\int \frac{x}{\sqrt{1-x^2}}u\sqrt{1-x^2}du=-2\int xu du$$

But how do I get rid of the $x$?

• $u = \arcsin x \implies x = \sin u$ Commented Sep 10, 2016 at 6:45
• $x=\sin u$! Hope that helps. Commented Sep 10, 2016 at 6:45
• @DougM Ahhhhh right Commented Sep 10, 2016 at 6:45

Hint

$$\int^t \arcsin^2(x)\mathrm d x=\int^{\arcsin(t)}u^2\cos(u)\mathrm d u.$$ Double integration by part gives you the solution !

Recall the series representation $$\arcsin^2(x)=\frac12\sum_{n\geq1}\frac{4^nx^{2n}}{n^2{2n\choose n}}$$ With this, we have the integral $$\int\arcsin^2(x)dx=\int\frac12\sum_{n\geq1}\frac{4^nx^{2n}}{n^2{2n\choose n}}dx$$ $$\int\arcsin^2(x)dx=\frac12\sum_{n\geq1}\frac{4^n}{n^2{2n\choose n}}\int x^{2n}dx$$ $$\int\arcsin^2(x)dx=\frac12\sum_{n\geq1}\frac{4^n}{n^2{2n\choose n}}\frac{x^{2n+1}}{2n+1}$$ And of course add your arbitrary constant in for good luck.