Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx},\text{ n}\in\text{N}$

Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)}{\sin^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)+\cos^{\text{n}}\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\cos^\text{n}\text{x}+\sin^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg[\frac{\sin^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}+\frac{\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\bigg]\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}{\sin^\text{n}\text{x}+\cos^\text{n}\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\big[\text{x}\big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$

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