Question

${\int }_0^{\frac{\pi }{2n}}\frac{dx}{1+{\cot }^{n}nx}=$.

Answer 1

Let $I={\int }_0^{\frac{\pi }{2n}}\frac{dx}{1+{\cot }^{n}nx}$

$\Rightarrow I={\int }_0^{\frac{\pi }{2n}}\frac{dx}{1+{\cot }^{n}n(\frac{\pi }{2n}-x)}={\int }_0^{\frac{\pi }{2n}}\frac{dx}{1+{\cot }^n(\frac{\pi }{2}-nx)}$

$\Rightarrow I={\int }_0^{\frac{\pi }{2n}}\frac{dx}{1+{\tan }^{n}nx}={\int }_0^{\frac{\pi }{2n}}\frac{dx}{1+\frac{1}{{\cot }^{n}nx}}={\int }_0^{\frac{\pi }{2n}}\frac{{\cot }^{n}nxdx}{{\cot }^{n}nx+1}$

Therefore, $2I={\int }_0^{\frac{\pi }{2n}}\frac{dx}{1+{\cot }^{n}nx}+{\int }_0^{\frac{\pi }{2n}}\frac{{\cot }^{n}nxdx}{{\cot }^{n}nx+1}={\int }_0^{\frac{\pi }{2n}}\frac{(1+{\cot }^{n}nx)dx}{1+{\cot }^{n}nx}={\int }_0^{\frac{\pi }{2n}}dx=\frac{\pi }{2n}$

So, $I=\frac{\pi }{4n}$