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All hell to integration

  1. Dec 1, 2003 #1
    OK. I've had a little break from my studdies and need some help with this...

    [itex]I_n(x)=\int\limits_0^x \tan^n{{\theta}}{{d\theta}},n\leq{0},{{x}}<\frac{\pi}{2}[\latex]

    By writing [itex]\tan{\theta}[\latex] as [itex]\tan^{n-2}{\theta}\tan^2{\theta}[\latex], or otherwise, show that

    [itex]I_n(x)=\frac{1}{n-1}\tan^{n-1}{x}-I_{n-2}(x), n\leq{2},x<\frac{\pi}{2}[\latex]

    Hence evaluate [itex]\int\limits_{0}^{\frac{\pi}{3}}\tan^4{\theta}d\theta[\latex], leaving your answers in terms of [itex]\pi[\latex]

    Thanks (Goddam further maths)

    AHHHH some one edit my post and get this bloody tex to work!!! pls
     
    Last edited by a moderator: Dec 1, 2003
  2. jcsd
  3. Dec 1, 2003 #2
    [tex]I_n(x)=\int\limits_0^x \tan^n \theta\,d\theta, n\leq{0}, x<\frac{\pi}{2}[/tex]

    By writing [itex]\tan{\theta}[/itex] as [itex]\tan^{n-2}{\theta}\tan^2{\theta}[/itex], or otherwise, show that

    [tex]I_n(x)=\frac{1}{n-1}\tan^{n-1}{x}-I_{n-2}(x), n\leq{2},x<\frac{\pi}{2}[/tex]

    Hence evaluate [itex]\int_0^{\pi/3}\tan^4{\theta}\,d\theta[/itex], leaving your answers in terms of [itex]\pi[/itex]
     
    Last edited: Dec 1, 2003
  4. Dec 1, 2003 #3

    Hurkyl

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    replace [ latex ] with [ tex ] (no spaces)
     
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