$\int_{0}^{\frac{\pi}{4}}{\frac{\cos(2x)}{\cos x+\sin x}dx}$

Question

$$\int_{0}^{\frac{\pi}{4}}{\frac{\cos(2x)}{\cos x+\sin x}dx}$$

I don't know how to evaluate this integral. This is a path that I've tried:

$$\int_{0}^{\frac{\pi}{4}}{\frac{\cos(2x)}{\cos x+\sin x}dx} =$$ $$= \int_{0}^{\frac{\pi}{4}}{\frac{\cos^2x - \sin^2x}{\cos x+\sin x}dx} =$$ $$= \int_{0}^{\frac{\pi}{4}}{\frac{1-2\sin^2x}{\cos x+\sin x}dx}$$

My idea was: let $t = \sin x$. Then $dt = \cos x dx$. But I only have $\frac{1}{\cos x}$. Any hints as to how to proceed from there?

This integral is giving me a hard time.

Answer 1

Hint: Notice, that:

$$\int_0^{\frac{\pi}{4}}\frac{\cos^2x - \sin^2x}{\cos{x}+\sin{x}}dx= \int_0^{\frac{\pi}{4}}\frac{(\cos{x} - \sin{x})(\cos{x}+\sin{x})}{\cos{x}+\sin{x}}dx= \int_0^{\frac{\pi}{4}}\left(\cos{x}-\sin{x}\right)dx$$