Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{1}{1+{\cot }^{7}x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let},I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+co{t}^{7}x}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+co{t}^7\left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+{\tan }^{7}x}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{1+co{t}^{7}x}+\frac{1}{1+{\tan }^{7}x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2+co{t}^{7}x+{\tan }^{7}x}{\left(1+co{t}^{7}x\right)\left(1+{\tan }^{7}x\right)}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2+co{t}^{7}x+{\tan }^{7}x}{2+co{t}^{7}x+{\tan }^{7}x}dx \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}dx \\ ={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}} \\ =\frac{\mathrm{\pi }}{2} \\ \mathrm{Hence},I=\frac{\mathrm{\pi }}{4} \end{gathered}$$