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Thread: Integration---I'm stuck!

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

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

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

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

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

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


    Therefore

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


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


    So the integral becomes

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

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

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