Prove $\int ^{\pi/2}_0 \frac{\sin^m x}{\sin^m x+\cos^m x}dx=\frac{\pi}{4}$

Question

Prove $$\int ^{\pi/2}_0 \frac{\sin^m x}{\sin^m x+\cos^m x}dx=\frac{\pi}{4}$$ for all real values of $m$.

Answer 1

Let $$I=\int ^{\pi/2}_0 \frac{\sin^m x}{\sin^m x+\cos^m x}dx$$ Let $$y=\frac{\pi}{2}-x$$ So $$I=\int ^{0}_{\pi/2} \frac{\sin^m (\frac{\pi}{2}-y)}{\sin^m (\frac{\pi}{2}-y)+\cos^m (\frac{\pi}{2}-y)}(-dy)=\int ^{\pi/2}_0 \frac{\cos^m y}{\cos^m y+\sin^m y}dy$$ So $$I+I=\int ^{\pi/2}_0 \frac{\sin^m x+\cos^m x}{\sin^m x+\cos^m x}dx=\frac{\pi}{2}$$ So $$I=\frac{\pi}{4}$$

Answer 2

First, check this fact

exercise:

If $f(x)$ is continuous on $[0,1]$, then

$$ \int\limits_0^{\frac{ \pi }{2}} f( \sin x) dx = \int\limits_0^{\frac{ \pi }{2}} f( \cos x) dx $$ To solve this, just use the change of variable $y = \pi/2 -x $ and compute.

Using the exercise, we have

$$ I = \int\limits_0^{\pi/2} \frac{ \sin^nx }{\sin^nx + \cos^n x } = \int\limits_0^{\pi/2} \frac{ \cos^nx }{\cos^nx + \sin^n x } $$

Thus,

$$ 2I = \int\limits_0^{\pi/2} dx = \frac{ \pi }{2} \implies \boxed{ I = \frac{ \pi }{4} } $$