How do I evaluate this integral :$\int \frac{1}{ \sqrt{1-x^2} } \arctan( \frac{\sqrt{1-x^2}}{2})dx$? Is there somone who can show me how do I evaluate this integral :$$\int \frac{1}{ \sqrt{1-x^2} } \arctan\left(\frac{\sqrt{1-x^2}}{2}\right)dx$$
Note: I took : $x=\cos t$ , but it didn't work 
Thank you for any help .
 A: We have:
$$\begin{eqnarray*} \int_{0}^{u}\arctan\left(\frac{\cos\theta}{2}\right)\,d\theta &=& \int_{0}^{u}\int_{0}^{\frac{\cos\theta}{2}}\frac{1}{1+t^2}\,dt\,d\theta\\&=&\int_{0}^{u}2\cos\theta\int_{0}^{1}\frac{1}{4+z^2\cos^2\theta}\,dz\,d\theta\\&=&\int_{0}^{\tan u}\frac{1}{\sqrt{1+v^2}}\int_{0}^{1/2}\frac{1}{1+v^2+z^2}\,dz\,dv\\&=&\int_{0}^{\text{arcsinh}\tan u}\int_{0}^{1/2}\frac{1}{z^2+\cosh^2\rho}\,dz\,d\rho\\&=&\int_{0}^{1/2}\text{arctanh}\left(\frac{z\sin u}{\sqrt{1+z^2}}\right)\frac{dz}{z\sqrt{1+z^2}}\\&=&\int_{2}^{+\infty}\text{arctanh}\left(\frac{\sin u}{\sqrt{1+w^2}}\right)\frac{dw}{\sqrt{1+w^2}}\\&=&\int_{\text{arcsinh} 2}^{+\infty}\text{arctanh}\left(\frac{\sin u}{\cosh \eta}\right)\,d\eta\\&=&\color{red}{\int_{2+\sqrt{5}}^{+\infty}\text{arctanh}\left(\frac{2\tau}{1+\tau^2}\sin u\right)\frac{d\tau}{\tau}}\end{eqnarray*}$$
hence the answer is given by a combination of dilogarithms, since $\text{arctanh}(x)=\frac{1}{2}\left(\log(1+x)-\log(1-x)\right)$ and:

$$ \int\log(1+\tau^2)\frac{d\tau}{\tau}=-\frac{1}{2}\text{Li}_2(-\tau^2),\qquad \int\log(1+\alpha\tau)\frac{d\tau}{\tau}= -\text{Li}_2(-\alpha\tau).$$

A: For the integral
\begin{align}
I = \int \frac{1}{ \sqrt{1-x^2} } \, \arctan\left(\frac{\sqrt{1-x^2}}{2}\right) \, dx
\end{align}
let $u^{2} = 1 - x^{2}$ to obtain
\begin{align}
I = - \, \int \tan^{-1}\left(\frac{u}{2}\right) \, \frac{du}{\sqrt{1-u^{2}}}.
\end{align}
Using $2 \tan^{-1}(x) = \ln(1 - ix) - \ln(1+ix)$ then after the transformation $2 t = i u$ the integral becomes
\begin{align}
I = i \, \int \ln\left( \frac{1-t}{1+t} \right) \, \frac{dt}{\sqrt{1 + 4 t^{2}}}. 
\end{align}
Making use of integration by parts where 
\begin{align}
u = \ln\left(\frac{1-x}{1+x}\right) \hspace{10mm} dv = \frac{dt}{\sqrt{1 + 4 t^{2}}}
\end{align}
to obtain
\begin{align}
-i \, I = \frac{1}{2} \, \ln\left(\frac{1-t}{1+t}\right) \, \sinh^{-1}(2t) - \int \frac{\sinh^{-1}(2t) \, dt}{t^{2} -1}.
\end{align}
From here use Wolfram Alpha to reduce the necessary terms. 
