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Math Help - cal II intergral problem

  1. #1
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    cal II intergral problem

    am running into trouble intergrating
    the integral tan^5(x) dx


    i know i need to break it up in terms of sec(x)^2 and tan(x)sec(x) so that i can use u subsitution




    also would you guys give me any tips on how to do well in cal II am taking a summer course right now
    and i need all the tips that i can get thank you very much
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  2. #2
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    Quote Originally Posted by momo55 View Post
    am running into trouble intergrating
    the integral tan^5(x) dx


    i know i need to break it up in terms of sec(x)^2 and tan(x)sec(x) so that i can use u subsitution




    also would you guys give me any tips on how to do well in cal II am taking a summer course right now
    and i need all the tips that i can get thank you very much
    \tan^5{x} = \tan^3{x}\tan^2{x}

     = \tan^3{x}(\sec^2{x} - 1)

     = \tan^3{x}\sec^2{x} - \tan^3{x}

     = \tan^3{x}\sec^2{x} - \tan{x}(\sec^2{x} - 1)

     = \tan^3{x}\sec^2{x} - \tan{x}\sec^2{x} + \tan{x}

     = \sec^2{x}(\tan^3{x} - \tan{x}) + \tan{x}.



    So \int{\tan^5{x}\,dx} = \int{\sec^2{x}(\tan^3{x} - \tan{x}) + \tan{x}\,dx}

     = \int{\sec^2{x}(\tan^3{x} - \tan{x})\,dx} - \int{\frac{-\sin{x}}{\phantom{-}\cos{x}}\,dx}.


    Now let u = \tan{x} so that du = \sec^2{x}\,dx and let v = \cos{x} so that dv = -\sin{x}\,dx, and the integrals become


    \int{u^3 - u\,du} - \int{\frac{1}{v}\,dv}

     = \frac{1}{4}u^4 - \frac{1}{2}u^2 - \ln{|v|} + C

     = \frac{1}{4}\tan^4{x} - \frac{1}{2}\tan^2{x} - \ln{|\cos{x}|} +C.
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  3. #3
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    the answer is
    sec(x)^4/4 - tan(x)^2/2 - ln|sec(x)|

    i just don't know how to get there
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  4. #4
    MHF Contributor matheagle's Avatar
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    IF you use the identity with tangent and secant, that was used to integrate your problem,
    you will be able to show that the two answers are within a constant (C), of being the same.

    However, one of these seem off, with the constant in front of the logarithm.

    \ln |\sec x|=-\ln |\cos x|
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  5. #5
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    Quote Originally Posted by momo55 View Post
    the answer is
    sec(x)^4/4 - tan(x)^2/2 - ln|sec(x)|

    i just don't know how to get there
    It should be a + in front of the logarithm there...

    -\ln{|\cos{x}|} = -\ln{\left|\frac{1}{\sec{x}}\right|}

     = -\ln{|\sec{x}|^{-1}}

     = \ln{|\sec{x}|}.
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  6. #6
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    yea i want to know the steps to get there
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  7. #7
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    Quote Originally Posted by momo55 View Post
    yea i want to know the steps to get there
    Read my posts...
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  8. #8
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    i mean understand exept for the tanx^4 and the ln|cosx| i donno if it makes much diffrence that in the book it says secx^4 and ln|secx| insted of tanx^4 and the ln|cosx|
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