- #1
Blandongstein
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I found this question on a website.
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.
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.