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Integral involving trigonometric functions.

I found this question on a website.

1. Homework Statement
Prove that [itex]\displaystyle \int_{0}^{\pi}\frac{1-\cos(nx)}{1-\cos(x)} dx=n\pi \ \ , n \in \mathbb{N}[/itex]

2. The attempt at a solution

Here's my attempt using induction:

Let [itex]P(n) [/itex] be the statement given by

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

P(1):

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

P(1) holds true.

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

[tex]\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[/tex]

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.
 
8
0
Let [itex]P(n)=\int\frac{1-\cos{nx}}{1-\cos{x}}dx[/itex]

[itex]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[/itex]
(thr limits of integral are 0 and pi)
The last Integral is the kernel of Dirighle and equal to [itex]\pi[/itex].
So [itex]P(n+1)-P(n)=\pi[/itex] .
Finally we obtain [itex]P(n)=n\pi[/itex].
 
Last edited:

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