Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{1+\tan x}dx$ is equal to

(a) $\frac{\mathrm{\pi }}{4}$

(b) $\frac{\mathrm{\pi }}{3}$

(c) $\frac{\mathrm{\pi }}{2}$

(d) π

Answer 1

(a) $\frac{\mathrm{\pi }}{4}$


$$\begin{gathered} \mathrm{Let},I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\tan x}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\tan \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+cotx}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get} \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{1+\tan \mathrm{x}}+\frac{1}{1+\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}\right]\mathit{d}x \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{\left(1+\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}\right)+\left(1+\tan \mathrm{x}\right)}{\left(1+\tan x\right)\left(1+\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}\right)}\right]\mathit{d}x \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{2+\tan \mathrm{x}+\cot \mathrm{x}}{1+\tan \mathrm{x}+\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}+\tan \mathrm{x}\cot \mathrm{x}}\right]\mathit{d}x \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{2+\tan \mathrm{x}+\cot \mathrm{x}}{2+\tan \mathrm{x}+\cot \mathrm{x}}\right]\mathit{d}x \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}}=\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence},I=\frac{\mathrm{\pi }}{4} \end{gathered}$$