How to prove this trigonometric integral? $$ \displaystyle \int_{-\pi/4}^{\pi/4} {{\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)}^{\cos(2t)} \ dx} = \frac{\pi}{2 \sin(\pi \cos^2 t)}$$
I could simplify it to 
$\displaystyle \int_0^1 {\left(t^n + \frac{1}{t^n}\right) \ \frac{dt}{1+t^2}}, \ n = \cos 2t $
From here, I can think of expanding into sums but that doesn't seem a good option. Also, getting back to trigonometric form is also an option but it would get us to reduction formula which will be messy. 
What is a straight, neat and easy approach to solve it?

\begin{align*}
  \frac{\cos x-\sin x}{\cos x+\sin x} &=
  \tan \left( \frac{\pi}{4}-x \right) \\
  \int_{-\pi/4}^{\pi/4}
  \left( \frac{\cos x-\sin x}{\cos x+\sin x} \right)^{n} dx &=
  \int_{-\pi/4}^{\pi/4} \tan^{n} \left( \frac{\pi}{4}-x \right) dx \\
  &= \int_{0}^{\pi/2} \tan^{n} u \, du \\
  &= \frac{\pi}{2} \sec \frac{n\pi}{2} \, , \quad -1<n<1 \\
  &= \frac{\pi}{2} \sec \frac{\pi \cos 2t}{2} \\
  &= \frac{\pi}{2} \sec \frac{\pi (2\cos^2 t-1)}{2} \\
  &= \frac{\pi}{2} \csc (\pi \cos^2 t) \\
\end{align*}
