Question

The value of the integral ${\int }_0^{\pi }\frac{\sin kx}{\sin x}dx$ $(k$ is an even integer) is equal to

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

Answer 1

The correct option is D 0

${\int }_0^{\pi }\frac{sinkx}{sinx}dx$
$sin(2n\pi -x)=-sinx$
$I={\int }_0^{\pi }\frac{sin(k\pi -kx)}{sin(\pi -x)}dx$
$=-{\int }_0^{\pi }\frac{sinkx}{sinx}dx$
$2I=0$
$I=0$