Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{1+\cot x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+cotx}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+cot\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}dx\left[\mathrm{Using}{\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\tan x}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+cotx}+\frac{1}{1+\tan x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1+\tan x+1+cotx}{\left(1+cotx\right)\left(1+\tan x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2+\tan x+cotx}{1+\tan x+cotx+\tan xcotx}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2+\tan x+cotx}{2+\tan x+cotx}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}}=\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence},I=\frac{\mathrm{\pi }}{4} \end{gathered}$$