Question

$\underset{0}{\overset{\pi /2}{\int }}\left(\frac{\sqrt[n]{secx}}{\sqrt[n]{secx}+\sqrt[n]{cosecx}}\right)dx=$

• A. $\pi /2$
• B. $\pi /3$
• C. $\pi /4$
• D. $\pi /6$

Answer 1

Answer: (C)

Let $I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x}+\sqrt[n]{cosecx}}dx\dots (i)$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sqrt[n]{\sec \left(\frac{\pi }{2}-x\right)}}{\sqrt[n]{\sec \left(\frac{\pi }{2}-x\right)}+\sqrt[n]{\mathrm{cosec}\left(\frac{\pi }{2}-x\right)}}dx$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sqrt[n]{\mathrm{cosec}x}}{\sqrt[n]{cosecx}+\sqrt[n]{\sec x}}dx\dots (ii)$
On adding Eq. (i) and (ii), we get
$2l=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sqrt[n]{\sec x}+\sqrt[n]{\mathrm{cosec}x}}{\sqrt[n]{\sec x}+\sqrt[n]{\mathrm{cosec}x}}dx$
$\Rightarrow 2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}dx=[x{]}_0^{\pi /2}$
$\Rightarrow 2I=\frac{\pi }{2}$
$\Rightarrow I=\frac{\pi }{4}$