# 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?

• Can you use the residue theorem? – jim May 7 '16 at 8:29

## 1 Answer

\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*}