Problem with $\int_{-\pi/2}^{\pi/2} \frac{\cos^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)}\ dx$

Question

Consider $S = \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\cos^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx$. We want to evaluate it.

I've tried :

$$\displaystyle \int_{-\pi/2}^{\pi/2} \frac{\cos^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx= \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\cos^{2n}(x) + \sin^{2n}(x) - \sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx =\\ \pi - \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx$$Hence we have $$\displaystyle \int_{-\pi/2}^{\pi/2} \frac{\cos^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx+ \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx = \pi \\ $$ Now I think that two left integrals are equal.

Let's try to prove it : $$\displaystyle \int_{-\pi/2}^{\pi/2} \frac{\cos^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx - \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx = \displaystyle \int_{0}^{\pi} \frac{\sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx - \displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx =\\ \displaystyle \int_{\pi/2}^{\pi} \frac{\sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx - \displaystyle \int_{-\pi/2}^{0} \frac{\sin^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)} dx = 0$$. Hence we have that $S = \frac{\pi}{2}$. Am I right?

Answer 1

$$S=\int_{-\pi/2}^{\pi/2} \frac{\cos^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)}\ dx=2\int_{0}^{\pi/2} \frac{\cos^{2n}(x)}{\sin^{2n}(x) + \cos^{2n}(x)}\ dx$$

let $\frac{\pi}{2}-x\to x$ we get

$$S=2\int_{0}^{\pi/2} \frac{\sin^{2n}(x)}{\cos^{2n}(x) + \sin^{2n}(x)}\ dx$$

now add the integral to both sides

$$\Longrightarrow 2S=2\int_{0}^{\pi/2} \frac{\sin^{2n}(x)+\cos^{2n}(x)}{\cos^{2n}(x) + \sin^{2n}(x)}\ dx=2\int_0^{\pi/2}\ dx=\pi \Longrightarrow S=\frac{\pi}{2}$$