Question

$\underset{\mathrm{\pi }/4}{\overset{\mathrm{\pi }/2}{\int }}\cot xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\cot x\mathit{}\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I=-{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\cot x\frac{-(\mathrm{cosec}x+\cot x)}{\mathrm{cosec}x+\cot x}dx \\ \Rightarrow I=-{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{-\mathrm{cosec}x\cot x-{\cot }^{2}x}{\mathrm{cosec}x+\cot x}dx \\ \Rightarrow I=-{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{-\mathrm{cosec}x\cot x-{\mathrm{cosec}}^{2}x+1}{\mathrm{cosec}x+\cot x}dx\left[\because {\mathrm{cosec}}^{2}x=1+{\cot }^{2}x\right] \\ \Rightarrow I=-{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{-\mathrm{cosec}x\cot x-{\mathrm{cosec}}^{2}x}{\mathrm{cosec}x+\cot x}dx-{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{1}{\mathrm{cosec}x+\cot x}dx \\ \Rightarrow I=-{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{-\mathrm{cosec}x\cot x-{\mathrm{cosec}}^{2}x}{\mathrm{cosec}x+\cot x}dx-{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}\frac{\sin x}{1+\cos x}dx \\ \Rightarrow I=-{\left[\log \left(\mathrm{cosec}x+\cot x\right)\right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}}+{\left[\log \left(1+\cos x\right)\right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}} \\ \Rightarrow I=-\log \left(1+\infty \right)+\log \left(\sqrt{2}+1\right)+\log \left(1+0\right)-\log \left(1+\frac{1}{\sqrt{2}}\right) \\ \Rightarrow I=\log \left(\sqrt{2}+1\right)-\log \left(\frac{\sqrt{2}+1}{\sqrt{2}}\right) \\ \Rightarrow I=\log \left(\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)}\right) \\ \Rightarrow I=\log \sqrt{2} \\ \Rightarrow I=\frac{1}{2}\log 2 \end{gathered}$$