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Math Help - Please help me solve this problem (Integration of Trig. function).

  1. #1
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    Smile Please help me solve this problem (Integration of Trig. function).



    I'm very new for calculus.
    please solve this problem. thank you very much.
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  2. #2
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    Quote Originally Posted by ronaldkaka View Post


    I'm very new for calculus.
    please solve this problem. thank you very much.

    Multiply top and bottom by ( sin^7 x - cos ^7 x)

    you get on the bottom sin^14 x - cos^14 x

    cos^2 x = 1 -sin^2x

    Use this identity then finish of the question.
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  3. #3
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    Thumbs up

    Thank you very much Mr.╔(σ_σ)╝.
    But please describe more.
    Last edited by ronaldkaka; May 7th 2009 at 06:02 PM. Reason: edit
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  4. #4
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    Quote Originally Posted by ronaldkaka View Post
    Thank you very much Mr.╔(σ_σ)╝.
    But please describe more.
    I gave you a useless step, sorry i wasn't thinking about it.

    Anyway i'm going to suggest something radical and an easy way out.

    But i'm not sure if you are allowed to leave your answers in terms of non elementary fuctions.


    Divide everything by cos^7 x

    you get \frac{tan^{7} x}{ 1 + tan^{7} x}  = 1 - \frac{1}{1 + tan^{7}x}


    use a substitution u = tanx

    du = sec^2 x = ( u^2 - 1)

    you get

    \frac{1}{ (u^{2} -1)( 1+u^{7})} you can do this

    \frac{1}{(u-1)(u+1)(1+u^{7})}

    Partial fractions would be appropriate.

    If you get something like

     \frac{Au+b}{(1+u^{7})}

    You may have to use the maclaurin series or do some more algrabra. I didn't try it so i'm not sure what you would end up with.

    Good luck though. The algebra seems disgusting.

    I popped it into an online integrator and it doesn't look pretty.
    Wolfram Mathematica Online Integrator
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  5. #5
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    Thumbs up

    Great.
    Thank you very much ╔(σ_σ)╝.
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  6. #6
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    I think you might get stuck with the partial fraction.

    \frac{1}{(u^{2}-1)(u^{7} +1)} =


    \frac{-3x^{4}+2x^{3} -4x^{2} +2x -3}{7( x^{6}-x^{5}+x^{4} -x^{3} +x^{2}-x +1)}   + \frac{1}{4(x-1)}- \frac{1}{4(x+1)} - \frac{1}{14(x+1)^{2}}

    Did it with pc.

    Forget about this integral; it's not the kind of thing you would see in an exam.
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  7. #7
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    Quote Originally Posted by ronaldkaka View Post


    I'm very new for calculus.
    please solve this problem. thank you very much.
    it's a very standard integral: do the substitution x \to \frac{\pi}{2} - x and then add the two integrals together to get \int_0^{\frac{\pi}{2}} dx=\frac{\pi}{2}. so the value of the integral is \frac{\pi}{4}.
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