Question

Solve $\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{dx}{1+{\cot }^{n}nx}$

• A. $0$
• B. $\frac{\pi }{4n}$
• C. $\frac{\pi }{2n}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $\frac{\pi }{2n}$

$\begin{array}{c}I=\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{dx}{1+{\cot }^{n}nx}\\ =\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{dx}{1+{\cot }^{n}n\left(\frac{\pi }{2n}-x\right)}\left[\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f\left(a-x\right)dx\right]\\ =\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{dx}{1+{\cot }^n\left(\frac{\pi }{2}-nx\right)}\\ =\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{dx}{1+{\tan }^{n}nx}\\ =\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{dx}{1+\frac{1}{{\cot }^{n}nx}}\\ =\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{{\cot }^{n}nx}{{\cot }^{n}nx+1}dx\\ \therefore 2I=\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{dx}{1+{\cot }^{n}nx}+\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{{\cot }^{n}nx}{{\cot }^{n}nx+1}dx\\ =\underset{0}{\overset{\frac{\pi }{2n}}{\int }}\frac{1+{\cot }^{n}nx}{{\cot }^{n}nx+1}dx\\ =\underset{0}{\overset{\frac{\pi }{2n}}{\int }}dx\\ ={\left[x\right]}_0^{\frac{\pi }{2n}}=\frac{\pi }{2n}\end{array}$