Question

$\underset{\mathrm{\pi }/6}{\overset{\mathrm{\pi }/3}{\int }}\frac{1}{1+\sqrt{\cot }x}dx$ is

(a) π/3
(b) π/6
(c) π/12
(d) π/2

Answer 1

$$\begin{gathered} \left(\mathrm{c}\right)\frac{\mathrm{\pi }}{12} \\ \mathrm{Let},I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+\sqrt{cotx}}dx...\left(\mathrm{i}\right) \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+\sqrt{cot\left({\displaystyle \frac{\mathrm{\pi }}{3}}+{\displaystyle \frac{\mathrm{\pi }}{6}}-x\right)}}dx\left[\mathrm{Using}{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right] \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+\sqrt{\tan x}}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get} \\ 2I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\left[\frac{1}{1+\sqrt{cotx}}+\frac{1}{1+\sqrt{\tan x}}\right]dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{2+\sqrt{cotx}+\sqrt{\tan x}}{\left(1+\sqrt{cotx}\right)\left(1+\sqrt{\tan x}\right)}dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\left[\frac{2+\sqrt{cotx}+\sqrt{\tan x}}{2+\sqrt{cotx}+\sqrt{\tan x}}\right]dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}dx \\ ={\left[x\right]}_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}} \\ =\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6} \\ =\frac{\mathrm{\pi }}{6} \\ \mathrm{Hence},I=\frac{\mathrm{\pi }}{12} \end{gathered}$$