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Integral involving trigonometric functions.

  1. Jan 20, 2012 #1
    I found this question on a website.

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


    [tex] \int_{0}^{\pi}\frac{1-\cos(x)}{1-\cos(x)} dx=\int_{0}^{\pi}dx=\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.
  2. jcsd
  3. Jan 21, 2012 #2
    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: Jan 21, 2012
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