The value of the integral $\int_0^π\frac{\sin(n+\frac{1}{2})x}{\sin x/2}dx(n ∈ N)$ is

$π$

Let $I_n =\int_0^π\frac{\sin(n+\frac{1}{2})x}{\sin x/2}dx$

$∴I_n-I_{n-1}=\int_0^π\frac{\left\{\sin(n+\frac{1}{2})x-\sin(n-\frac{1}{2})x\right\}}{\sin(x/2)}dx$

$=\int_0^π\frac{2\cos nx.\sin(x/ 2)}{\sin(x / 2)}dx=\int_0^π2\cos nx\,dx$

$=2\left\{\frac{\sin nx}{n}\right\}_0^π= 0 – 0 = 0⇒I_n = I_{n–1}$

Replacing n by n – 1, n – 2, ..., then we get

$I_n = I_{n–1} = I_{n–2} = .... = I_1$

$⇒I_n = I_1=\int_0^π\frac{\sin(3x / 2)}{\sin(x / 2)}dx=\int_0^π\left(\frac{\sin 2x+\sin x}{\sin x}\right)dx$

$=\int_0^π(2 \cos x +1) dx=\{2 \sin x+ x\}_0^π=π$

Hence, $I_n = π$