The value of the integral $\int\limits_0^{\pi / 2} \sin 2 n x \cot x d x$, where n is a positive integer, is
Answer & explanation
Correct answer: option 1
We have,
$\cos x+\cos 3 x+\cos 5 x+...+\cos (2 n-1) x$
$=\frac{\cos [x+(n-1) x] \sin n x}{\sin x}$
$\Rightarrow 2[\cos x+\cos 3 x+\cos 5 x+...+\cos (2 n-1) x]=\frac{\sin 2 n x}{\sin x}$
$\Rightarrow \sin 2 n x \cot x=2 \cos x\{\cos x+\cos 3 x+\cos 5 x +...+\cos (2 n-1) x\}$
$\Rightarrow \sin 2 n x \cot x=1+2\{\cos 2 x+\cos 4 x+\cos 6 x +...+\cos (2 n-2) x\}+\cos 2 n x$
$\Rightarrow \int\limits_0^{\pi / 2} \sin 2 n x \cot x d x =\frac{\pi}{2}+2 \int\limits_0^{\pi / 2}\{\cos 2 x+\cos 4 x+... +\cos (2 n-2) x\} d x+\int\limits_0^{\pi / 2} \cos 2 n x d x$
$\Rightarrow \int\limits_0^{\pi / 2} \sin 2 n x \cot x d x =\frac{\pi}{2}+2 \times 0+0=\frac{\pi}{2}$