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

Question

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

Answer 1

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}

Answer 2

Let $f(a)$ be given by the integral

$$f(a)=\int_0^{\pi/2}\frac{\arctan(a\tan(x))}{\tan(x)}\,dx$$

Then, differentiating reveals

$$f'(a)=\int_0^{\pi/2}\frac{1}{1+a^2\tan^2(x)}\,dx=\frac{\pi/2 }{a+1}$$

Integrating and using $f(0)=0$ yields

$$f(a)=\frac\pi2\log(a+1)$$

Therefore, we find that

$$\int_0^{\pi/2}\frac{x}{\tan(x)}\,dx=\frac\pi 2 \log(2)$$