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Q. $\int\limits_{0}^{\pi/2} \left(\frac{\sqrt[n]{sec\,x}}{\sqrt[n]{sec\,x} +\sqrt[n]{cosec\,x}}\right) dx=$

MHT CETMHT CET 2016Integrals

Solution:

Let $I=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x}+\sqrt[n]{cosec \,x}} d x \,\dots(i)$
$=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sqrt[n]{\sec \left(\frac{\pi}{2}-x\right)}}{\sqrt[n]{\sec \left(\frac{\pi}{2}-x\right)}+\sqrt[n]{\operatorname{cosec}\left(\frac{\pi}{2}-x\right)}} d x$
$=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sqrt[n]{\operatorname{cosec} x}}{\sqrt[n]{cosec \,x}+\sqrt[n]{\sec x}} d x\, \dots(ii)$
On adding Eq. (i) and (ii), we get
$2 l=\int\limits_{0}^{\frac{\pi}{2}} \frac{\sqrt[n]{\sec x}+\sqrt[n]{\operatorname{cosec} x}}{\sqrt[n]{\sec x}+\sqrt[n]{\operatorname{cosec} x}} d x$
$\Rightarrow 2I=\int\limits_{0}^{\frac{\pi}{2}} d x=[x]_{0}^{\pi / 2}$
$ \Rightarrow 2I=\frac{\pi}{2}$
$ \Rightarrow I=\frac{\pi}{4}$