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Question

# $\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{1+\mathrm{tan}x}dx$ is equal to (a) $\frac{\mathrm{\pi }}{4}$ (b) $\frac{\mathrm{\pi }}{3}$ (c) $\frac{\mathrm{\pi }}{2}$ (d) π

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Solution

## (a) $\frac{\mathrm{\pi }}{4}$ $\mathrm{Let},I={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\mathrm{tan}x}dx...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+\mathrm{tan}\left(\frac{\mathrm{\pi }}{2}-x\right)}dx\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+cotx}dx...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}2I={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\left[\frac{1}{1+\mathrm{tan}\mathrm{x}}+\frac{1}{1+\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}}\right]\mathit{d}x\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\left[\frac{\left(1+\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}\right)+\left(1+\mathrm{tan}\mathrm{x}\right)}{\left(1+\mathrm{tan}x\right)\left(1+\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}\right)}\right]\mathit{d}x\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\left[\frac{2+\mathrm{tan}\mathrm{x}+\mathrm{cot}\mathrm{x}}{1+\mathrm{tan}\mathrm{x}+\mathrm{c}\mathrm{o}\mathrm{t}\mathrm{x}+\mathrm{tan}\mathrm{x}\mathrm{cot}\mathrm{x}}\right]\mathit{d}x\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}\left[\frac{2+\mathrm{tan}\mathrm{x}+\mathrm{cot}\mathrm{x}}{2+\mathrm{tan}\mathrm{x}+\mathrm{cot}\mathrm{x}}\right]\mathit{d}x\phantom{\rule{0ex}{0ex}}={\int }_{0}^{\frac{\mathrm{\pi }}{2}}dx\phantom{\rule{0ex}{0ex}}={\left[x\right]}_{0}^{\frac{\mathrm{\pi }}{2}}=\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},I=\frac{\mathrm{\pi }}{4}$

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