If $n\in N$  then $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin 2nx \cot xdx=$ 

$I_n={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin 2nx\cot xdx}$

$I_n-I_{n-1}={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\sin 2nx-\sin 2\left(n-1\right)x\right]\cot xdx}$

                 $=2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \left(2n-1\right)x\cdot \cos xdx}$

                $={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left[\cos 2nx+\cos 2\left(n-1\right)x\right]dx}$

               $=\frac{\sin 2nx}{2n}{+\frac{\sin 2\left(n-1\right)x}{2\left(n-1\right)}|}_0^{\frac{\pi }{2}}=0$

$\therefore I_n=I_{n-1}=\mathrm{.......}=I_2=I_1=2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2}xdx=2\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{\pi }{2}}$