Derivating with respect the parameter calcule $\int_0^{\frac{\pi}{2}}\frac{\arctan(a\tan x)}{\tan x}dx$ Derivating with respect the parameter calcule $$\int_0^{\frac{\pi}{2}}\frac{\arctan(a\tan x)}{\tan x}dx$$

To complement the post by @MarkViola, let us show
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x}= \frac{\pi}{2(a+1)}
\end{align}
assuming $a>0$. 
Observe
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x} = \int^{\pi/2}_0 \frac{\sec^2 x- \tan^2x}{1+a^2\tan^2x}\ dx
\end{align}
which means
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x}+ \frac{1}{a^2}\int^{\pi/2}_0 \frac{a^2\tan^2 x}{1+a^2\tan^2x}\ dx = \int^{\pi/2}_0 \frac{\sec^2 x}{1+a^2\tan^2x}\ dx.
\end{align}
Next, notice the left hand side of the above identity is equal to
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x}+ \frac{1}{a^2}\int^{\pi/2}_0 1- \frac{1}{1+a^2\tan^2 x}\ dx = \left(1-\frac{1}{a^2} \right)\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2 x}+ \frac{\pi}{2a^2}.
\end{align}
and the right hand side by $u$-sub equals
\begin{align}
\int^{\pi/2}_0 \frac{\sec^2 x}{1+a^2\tan^2 x}\ dx = \frac{\pi}{2a}.
\end{align}
Hence it follows
\begin{align}
\left(1-\frac{1}{a^2} \right)\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2 x} = \frac{\pi(a-1)}{2 a^2} \ \ \ \implies \ \ \ \int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2 x} = \frac{\pi}{2(a+1)}.
\end{align}
Remark: If we drop the condition $a>0$, then we see that
\begin{align}
\int^{\pi/2}_0 \frac{\sec^2 x}{1+a^2\tan^2 x}\ dx = \frac{\pi}{2|a|}
\end{align}
which leads to the conclusion
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2 x} = \frac{\pi}{2(|a|+1)}.
\end{align}
