<s>[INST] 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 .
 [/INST] 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. 
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