Integral involving trigonometric functions.

1. Jan 20, 2012

Blandongstein

I found this question on a website.

1. The problem statement, all variables and given/known data
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.

2. Jan 21, 2012

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 $P(n)=n\pi$.

Last edited: Jan 21, 2012