Question

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx, n \in N .\]

Answer 1

\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin^n x}{\sin^n x + \cos^n x} d x ..................(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{\sin^n \left( \frac{\pi}{2} - x \right)}{\sin^n \left( \frac{\pi}{2} - x \right) + \cos^n \left( \frac{\pi}{2} - x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\cos^n x + \sin^n x} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\cos^n x}{\sin^n x + \cos^n x} d x ................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_0^\frac{\pi}{2} \left[ \frac{\sin^n x}{\sin^n x + \cos^n x} + \frac{\cos^n x}{\sin^n x + \cos^n x} \right] d x \]
\[ = \int_0^\frac{\pi}{2} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x} dx\]
\[ = \int_0^\frac{\pi}{2} dx\]
\[ = \left[ x \right]_0^\frac{\pi}{2} \]
\[ = \frac{\pi}{2}\]
\[Hence\ I = \frac{\pi}{4}\]