Integral involving trigonometric functions.

[Blandongstein]

I found this question on a website.

Homework Statement

Prove that \displaystyle \int_{0}^{\pi}\frac{1-\cos(nx)}{1-\cos(x)} dx=n\pi \ \ , n \in \mathbb{N}

2. The attempt at a solution

Here's my attempt using induction:

Let P(n) be the statement given by

\int_{0}^{\pi}\frac{1-\cos(nx)}{1-\cos(x)} dx=n\pi \ \ , n \in \mathbb{N}

P(1):

\int_{0}^{\pi}\frac{1-\cos(x)}{1-\cos(x)} dx=\int_{0}^{\pi}dx=\pi
(1)\pi=\pi

P(1) holds true.

Let P(n) be true.
Now, we need to show that P(n+1) is true.

\int_{0}^{\pi} \frac{1-\cos{x(n+1)}}{1-\cos(x)}dx=\int_{0}^{\pi}\frac{1-[\cos(nx)\cos(x)-\sin(nx)\sin(x)]}{1-\cos(x)} dx

I don't know how to proceed from here.
Furthermore, I would like to know if I could prove the statement by solving the integral.

 

[hrach87]

Let P(n)=\int\frac{1-\cos{nx}}{1-\cos{x}}dx

P(n+1)-P(n)=\int\frac{\cos{(n+1)x}-\cos{nx}}{1- \cos{x}}dx=\int\frac{2\sin{(n+\frac{1}{2})x} \sin{\frac{x}{2}}}{2\sin^{2}{\frac{x}{2}}}dx=\int \frac{\sin{(n+\frac{1}{2})x} }{\sin{\frac{x}{2}}}dx
(thr limits of integral are 0 and pi)
The last Integral is the kernel of Dirighle and equal to \pi.
So P(n+1)-P(n)=\pi .
Finally we obtain P(n)=n\pi.