# integral of rational trigonometric function involving arbitrary powers [duplicate]

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

I've tried dividing by $\cos^{m}(x)$, and subbing out the $\ 1+\cot^{m}(x)$ with $\csc^{n}(x)$ for some $n$, but to no avail. I've also tried adding and subtracting $\cos^{m}(x)$ to the numerator, and substituting $x$ by $\pi-y$, but these techniques haven't helped either.

• Trivial by symmetry and already asked many times. Jan 12, 2018 at 15:32

Let $I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{m}x}{\sin^{m}x+\cos^{m}x}dx$, by letting $y=\pi/2-x$, then $I=\displaystyle\int_{0}^{\pi/2}\dfrac{\cos^{m}y}{\sin^{m}y+\cos^{m}y}dy$, so $2I=\displaystyle\int_{0}^{\pi/2}\dfrac{\sin^{m}x+\cos^{m}x}{\sin^{m}x+\cos^{m}x}dx=\pi/2$, so $I=\pi/4$.
Let $$I=\int_{0}^{\pi /2} \frac{\sin^{m}(x)}{\sin^{m}(x)+\cos^{m}(x)} \, dx.$$ Then using the substitution $u=\frac{\pi}{2}-x$, we get $$I=\int_{0}^{\pi /2} \frac{\cos^{m}(u)}{\sin^{m}(u)+\cos^{m}(u)} \, du=\int_{0}^{\pi /2} \frac{\cos^{m}(x)}{\sin^{m}(x)+\cos^{m}(x)} \, dx.$$ Thus $$2I=\int_{0}^{\pi /2} \frac{\sin^{m}(x)+\cos^{m}(x)}{\sin^{m}(x)+\cos^{m}(x)} \, dx=\frac{\pi}{2}.$$ Hence $I=\frac{\pi}{4}$.