Practicing Success
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 |
$\frac{\pi}{2}$ $\frac{\pi}{4}$ $\frac{\pi}{2 n}$ $\frac{\pi}{4 n}$ |
$\frac{\pi}{4}$ |
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 $2 \pi=\int\limits_0^{\pi / 2} 1 . d x=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$ |