The value of $\int\limits_0^{\pi / 2 n} \frac{1}{1+\cot n x} d x$, is

0 $\frac{\pi}{4 n}$ $\frac{\pi}{2 n}$ $\frac{\pi}{2}$

$\frac{\pi}{4 n}$

Let $I=\int\limits_0^{\pi / 2 n} \frac{1}{1+\cot n x} d x=\int\limits_0^{\pi / 2 n} \frac{\sin n x}{\cos n x+\sin n x} d x$ Putting $n x=t$, we get $I=\frac{1}{n} \int\limits_0^{\pi / 2} \frac{\sin t}{\cos t+\sin t} d t=\frac{1}{n} \times \frac{\pi}{4}=\frac{\pi}{4 n}$