Question

If for $k\in N,$ $\frac{\sin 2kx}{\sin x}=2[\cos x+\cos 3x+...+\cos (2k-1\left)x\right],$
then value of $I={\int }_0^{\pi /2}\sin 2kx\cot xdx$ is

• A. $-\pi /2$
• B. $0$
• C. $\pi /2$
• D. $\pi$

Answer 1

Answer: (C)

$\pi /2$

Writing $I={\int }_0^{\pi /2}\frac{\sin 2kx}{\sin x}\cos xdx$
and using the given identity, we can write
$I={\int }_0^{\pi /2}2[\cos x+\cos 3x+......\cos (2k-1\left)x\right]\cos xdx$
$={\int }_0^{\pi /2}\left[\right(1+\cos 2x)+(\cos 4x+\cos 2x)+...+\cos 2kx+\cos (2k-2\left)x\right)]dx$
$={(x+\sin 2x+\frac{1}{2}\sin 4x+...+\frac{1}{2k}(\sin 2kx)]}_0^{\pi /2}$ $=\frac{\pi }{2}$