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Math Help - Integration---I'm stuck!

  1. #1
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    Integration---I'm stuck!

    \int \tan^4\theta d\theta
    I know you're supposed to use the identity \tan^2\theta+1=\sec^2\theta
    so, \int \tan^4\theta d\theta=\int (\sec^2\theta-1)^2 d\theta
    now I don't know what to do next. Can anyone give me some pointers to get moving?
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    \int \tan^4\theta d\theta
    I know you're supposed to use the identity \tan^2\theta+1=\sec^2\theta
    so, \int \tan^4\theta d\theta=\int (\sec^2\theta-1)^2 d\theta
    now I don't know what to do next. Can anyone give me some pointers to get moving?
    Thanks
    \tan^4{\theta} = \tan^2{\theta}\cdot\tan^2{\theta}

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

     = \tan^2{\theta}\sec^2{\theta} - \tan^2{\theta}

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

     = \tan^2{\theta}\sec^2{\theta} - \sec^2{\theta} + 1.


    Therefore

    \int{\tan^4{\theta}\,d\theta} = \int{\tan^2{\theta}\sec^2{\theta}\,d\theta} - \int{\sec^2{\theta}\,d\theta} + \int{1\,d\theta}


    Let u = \tan{\theta} so that \frac{du}{d\theta} = \sec^2{\theta}


    So the integral becomes

    \int{u^2\,du} - \int{\sec^2{\theta}\,d\theta} + \int{1\,d\theta}

    = \frac{1}{3}u^3 - \tan{\theta} + \theta + C

     = \frac{1}{3}\tan^3{\theta} - \tan{\theta} + \theta + C.
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