TITLE: Calculate $\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$
QUESTION [11 upvotes]: I am preparing for a calculus exam and I was asked to calculate $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx$$
Using the hint that $$\frac{\arctan(x)}{x}=\int_{0}^{1}\frac{dy}{1+x^2y^2}$$
I ran into some trouble and would appreciate help.
What I did:
I used the hint, $$\int_{0}^{1}\frac{\arctan(x)}{x\sqrt{1-x^2}}dx=\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2y^2)\sqrt{1-x^2}}dydx$$
Since $x,y$ move between $0,1$ I thought maybe it is best to used the transform $x=r\cos\theta$ , $y=r\sin\theta$. $r \in [0,1]$, $\theta \in [0, \frac{\pi}{2}]$
But I didn't get anything meaningful, I didn't end up an something that is easy / possible to integrate. And I honestly can't think of a way to integrate this as it is.

REPLY [8 votes]: May I show another way?. It does not use the given hint though.
As Norbert done, make the sub $\displaystyle x=\sin(t), \;\ dx=\cos(t)dt$
$$\int_{0}^{\frac{\pi}{2}}\frac{\tan^{-1}(\sin(t))}{\sin(t)}dt$$
Now, use the series for arctan: $\displaystyle \tan^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}$
only, $x=\sin(t)$ and we get:
$$\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}\frac{\sin^{2n}(t)(-1)^{n}}{2n+1}dt$$
But, using the famous and known result: $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}(t)dt=\frac{\pi (2n)!}{2^{2n+1}(n!)^{2}}$, one obtains the series:
$$\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(-1)^{n} (2n)!}{(2n+1)2^{2n}(n!)^{2}}$$
Now, notice what that series is?.  It is the Taylor series for $\displaystyle \sinh^{-1}(1)$
which is $$\sinh^{-1}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(2n)!x^{2n+1}}{2^{2n}(n!)^{2}(2n+1)}$$
so, we have $$\frac{\pi}{2}\sinh^{-1}(1)$$