Question

Evaluate the following integrals:

${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+{\cot }^{{\displaystyle \frac{3}{2}}}x}dx$

Answer 1

Let I = ${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+{\cot }^{{\displaystyle \frac{3}{2}}}x}dx$ .....(1)
Then,

$$\begin{gathered} I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+{\cot }^{{\displaystyle \frac{3}{2}}}\left({\displaystyle \frac{\mathrm{\pi }}{3}}+{\displaystyle \frac{\mathrm{\pi }}{6}}-x\right)}dx\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right] \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+{\cot }^{{\displaystyle \frac{3}{2}}}\left({\displaystyle \frac{\mathrm{\pi }}{2}-x}\right)}dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1}{1+{\tan }^{{\displaystyle \frac{3}{2}}}x}dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{{\cot }^{{\displaystyle \frac{3}{2}}}x}{{\cot }^{\frac{3}{2}}x+1}dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{1+{\cot }^{{\displaystyle \frac{3}{2}}}x}{1+{\cot }^{\frac{3}{2}}x}dx \\ \Rightarrow 2I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}dx \\ \Rightarrow 2I={\overline{)x}}_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}} \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{6} \\ \Rightarrow I=\frac{\mathrm{\pi }}{12} \end{gathered}$$