If $m, n \in N$, then $\int\limits_0^{\pi / 2} \frac{\sqrt[n]{\sin ^m x}}{\sqrt[n]{\sin ^m x}+\sqrt[n]{\cos ^m x}} d x$ is equal to
Answer & explanation
Correct answer: option 2
Let $I=\int\limits_0^{\pi / 2} \frac{\sqrt[n]{\sin ^m x}}{\sqrt[n]{\sin ^m x}+\sqrt[n]{\cos ^m x}} d x$ ....(i)
Then,
$I=\int\limits_0^{\pi / 2} \frac{\sqrt[n]{\cos ^m x}}{\sqrt[n]{\cos ^m x}+\sqrt[n]{\sin ^m x}} d x$ [Using $\int\limits_0^a f(x) dx = \int\limits_0^a f(a-x)dx$] ......(ii)
Adding (i) and (ii), we get
$2 \pi=\int\limits_0^{\pi / 2} 1 . d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$