$ I = \int_0^{\frac{\pi}{2}} \frac{\sin^n x}{\sin^n x + \cos^n x}dx $
$ x = \frac{\pi}{2} - u $
$ dx = -du $
$ I = -\int^0_{\frac{\pi}{2}} \frac{\sin^n (\frac{\pi}{2} - u)}{\sin^n (\frac{\pi}{2}-u) + \cos^n (\frac{\pi}{2}-u)}du = \int_0^{\frac{\pi}{2}} \frac{\sin^n (\frac{\pi}{2} - u)}{\sin^n (\frac{\pi}{2}-u) + \cos^n (\frac{\pi}{2}-u)}du = \int_0^{\frac{\pi}{2}} \frac{\cos^n u}{\cos^n u + \sin^n u}du $
$ 2I = \int_0^{\frac{\pi}{2}} \frac{\sin^n x}{\sin^n x + \cos^n x}dx + \int_0^{\frac{\pi}{2}} \frac{\cos^n x}{\sin^n x + \cos^n x}dx = \int_0^{\frac{\pi}{2}} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x}dx = \int_0^{\frac{\pi}{2}} dx = \frac{\pi}{2} $
$ I = \frac{\pi}{4} $
