Question

The value of ${\int }_0^{\pi }\frac{sin(n+\frac{1}{2})x}{sin\left(\frac{x}{2}\right)}dx$ is

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

Answer 1

The correct option is C $\pi$

We have
$2sin\frac{x}{2}cosx=sin\frac{3x}{2}-sin\frac{x}{2}$
$2sin\frac{x}{2}cos2x=sin\frac{5x}{2}-sin\frac{3x}{2}$
$\begin{array}{cccc}\cdots & \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots & \cdots \end{array}$
$2sin\frac{x}{2}cosnx=sin(n+\frac{1}{2})x-sin(n-\frac{1}{2})x$
Adding the above n equations we get
$2sin\frac{x}{2}(cosx+cos2x+\cdots +connx)$
$=sin(n+\frac{1}{2})x-sin\frac{x}{2}$
$\Rightarrow \frac{sin(n+\frac{1}{2})x}{sin\frac{x}{2}}-1=2(cosx+cos2x+\cdots cosnx)$
$\Rightarrow {\int }_0^{\pi }\frac{sin(n+\frac{1}{2})x}{sin\frac{x}{2}}dx-{\int }_0^{\pi }1dx=0$
$\Rightarrow {\int }_0^{\pi }\frac{sin(n+\frac{1}{2})x}{sin\frac{x}{2}}dx-\pi =0$
$\therefore {\int }_0^{\pi }\frac{sin(n+\frac{1}{2})x}{sin\frac{x}{2}}dx=\pi$