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Thread: a problem about integral

  1. #1
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    a problem about integral


    could you help me please. any help will be appreciated...
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: a problem about integral

    Have you tried with induction? (Maybe that will work)
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: a problem about integral

    Quote Originally Posted by mami View Post
    could you help me please. any help will be appreciated...
    Could you help us mentioning the subject where that integral is included? For example, sometimes we solve an integral using Complex Variable and we receive the following answer: Sorry, we have not covered Complex Variable, etc., etc., etc.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: a problem about integral

    I have become interested in this problem and have spent a great deal of time trying to get induction (among other things) to work, but have not been successful. Hopefully, someone can show me where I am going wrong. Obviously the statement is true for the base case of n = 1. So I stated the induction hypothesis P_n:

    \int_0^{\pi}\frac{\cos(nx)-\cos(na)}{\cos(x)-\cos(a)}\,dx=\pi\frac{\sin(na)}{\sin(a)}

    Assuming this is true, I then added P_{n+2} to get:

    \int_0^{\pi}\frac{\cos(nx)-\cos(na)}{\cos(x)-\cos(a)}\,dx+\int_0^{\pi}\frac{\cos\left((n+2)x \right)-\cos\left((n+2)a \right)}{\cos(x)-\cos(a)}\,dx=

    \pi\frac{\sin(na)}{\sin(a)}+\pi\frac{\sin\left((n+  2)a\right)}{\sin(a)}

    \int_0^{\pi}\frac{\left(\cos\left((n+2)x \right)+\cos(nx)\right)-\left(\cos\left((n+2)a\right)+\cos(na) \right)}{\cos(x)-\cos(a)}\,dx=\frac{\pi}{\sin(a)}\left(\sin\left((n  +2)a\right)+\sin(na)\right)

    \int_0^{\pi}\frac{\left(2\cos\left((n+1)x \right)\cos(x)\right)-\left(2\cos\left((n+1)a\right)\cos(a)\right)}{\cos  (x)-\cos(a)}\,dx=\frac{\pi}{\sin(a)}\(2\sin\left((n+1)  a\right)\cos(a)\)

    \int_0^{\pi}\frac{\cos\left((n+1)x\right) \frac{\cos(x)}{\cos(a)}-\cos\left((n+1)a\right)}{\cos(x)-\cos(a)}\,dx=\pi\frac{\sin\left((n+1)a \right)}{\sin(a)}

    I can't figure out how to get rid of the \frac{\cos(x)}{\cos(a)} in the numerator of the integrand.

    edit: sorry about my edits...this board has some unexpected idiosyncrasies concerning the
    \left and \right operators in \LaTeX...
    Last edited by MarkFL; Jan 2nd 2012 at 10:53 PM.
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: a problem about integral

    I should add that this problem has also been posted on the forum which I help moderate, and I have posted the same problem I have in using induction.

    The OP is an infrequent visitor there, and it may be a while before we hear from them to get any feedback on the context of the problem.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: a problem about integral

    The following fix was given to me:

    \int_0^{\pi}\frac{\cos((n+1)x)}{\cos(a)}\,dx=0

    \int_0^{\pi}\frac{\cos((n+1)x)\left(\frac{\cos(x)}  {\cos(a)}-1\right)+\cos((n+1)a)-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx=0

    \int_0^{\pi}\frac{\cos((n+1)x)\frac{\cos(x)}{\cos(  a)}-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx=\int_0^{\pi}\frac{\cos((n+1)x)-\cos((n+1)a)}{\cos(x)-\cos(a)}\,dx
    Last edited by MarkFL; Jan 4th 2012 at 09:54 AM.
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