calculus - Integral $int_0^{pi/4}frac{dx}{{sin 2x},(tan^ax+cot^ax)}=frac{pi}{8a}$

I am trying to prove this interesting integral $$\mathcal{I}:=\int_0^{\pi/4}\frac{dx}{{\sin 2x}\,(\tan^ax+\cot^ax)}=\frac{\pi}{8a},\qquad \mathcal{Re}(a)\neq 0.$$
This result is breath taking but I am more stumped than usual. It truly is magnificent. I am not sure how to approach this,
note $\sin 2x=2\sin x \cos x$. I am not sure how to approach this because of the term
$$(\tan^ax+\cot^ax)$$
in the denominator. I was trying to use the identity
$$\tan \left(\frac{\pi}{2}-x\right)=\cot x$$
since this method solves a similar kind of integral but didn't get anywhere. A bad idea I tried was to try and differentiate with respect to a
$$\frac{dI(a)}{da}=\int_0^{\pi/4}\partial_a \left(\frac{dx}{{\sin 2x}\,(\tan^ax+\cot^ax)}\right)=\int_0^{\pi/4} \frac{(\cot^a x \log(\cot x )+\log(\tan x ) \tan^a x)}{\sin 2x \, (\cot^a x+\tan^a x)^2}dx$$

which seems more complicated when I break it up into two integrals. How can we solve the integral? Thanks.

Answer

Rewrite:
$$\begin{align} \int_0^{\Large\frac\pi4}\frac{dx}{{\sin 2x}\,(\tan^ax+\cot^ax)}&=\int_0^{\Large\frac\pi4}\frac{dx}{{2\sin x\cos x}\,\left(\tan^ax+\dfrac1{\tan^ax}\right)}\\ &=\frac12\int_0^{\Large\frac\pi4}\frac{\tan^ax\ dx}{{\tan x\cos^2x}\,\left(1+\tan^{2a}x\right)}\\ &=\frac12\int_0^{\Large\frac\pi4}\frac{\tan^{a-1}x\ d(\tan x)}{1+\tan^{2a}x}. \end{align}$$
Now, let $\tan^a x=\tan\theta\;\Rightarrow\;a\tan^{a-1}x\ d(\tan x)=\sec^2\theta\ d\theta$. Then
$$\begin{align} \frac12\int_{x=0}^{\Large\frac\pi4}\frac{\tan^{a-1}x\ d(\tan x)}{1+\tan^{2a}x}&=\frac1{2|a|}\int_{\theta=0}^{\Large\frac\pi4}\frac{\sec^2\theta\ d\theta}{1+\tan^{2}\theta}\\ &=\frac1{2|a|}\int_{\theta=0}^{\Large\frac\pi4}\ d\theta\\ &=\large\color{blue}{\frac{\pi}{8|a|}}. \end{align}$$