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Math Help - Trigonometric Substitution Integrals

  1. #1
    Senior Member polymerase's Avatar
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    Trigonometric Substitution Integrals

    How do u integrate this function:

    <br />
\int x^2 {\sqrt{x^2-16}}\,dx\,<br />

    I got it down to this but now i cant solve this integral:

    <br />
256\int \tan^2\theta\,sec^3\theta\, d\theta\,<br />

    Which identity do i sub in....?
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  2. #2
    MHF Contributor red_dog's Avatar
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    \displaystyle\int\tan^2x\sec^3xdx=\int\frac{\sin^2  x}{\cos^5x}dx=\int\frac{\sin^2x\cos x}{\cos^6x}dx=\int\frac{\sin^2x}{(1-\sin^2x)^3}\cos xdx
    Now \sin x=t\Rightarrow\int\displaystyle\frac{t^2}{(1-t^2)^3}dt
    and this is a rational integral.
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by red_dog View Post
    \displaystyle\int\tan^2x\sec^3xdx=\int\frac{\sin^2  x}{\cos^5x}dx=\int\frac{\sin^2x\cos x}{\cos^6x}dx=\int\frac{\sin^2x}{(1-\sin^2x)^3}\cos xdx
    Now \sin x=t\Rightarrow\int\displaystyle\frac{t^2}{(1-t^2)^3}dt
    and this is a rational integral.
    How do u solve this function...it's almost as hard as the first
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  4. #4
    Eater of Worlds
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    I may be mistaken ,but I believe that's 128 and not 256.


    Anyway, \int{tan^{2}(x)sec^{3}(x)}dx

    Using identities, rewrite as:

    \int{(sec^{2}(x)-1)sec^{3}(x)}dx

    = \int{sec^{5}(x)}dx-\int{sec^{3}(x)}dx

    Now, you can use the reduction formula for sec:

    \int{sec^{m}(x)}dx=\frac{1}{m-1}sec^{m-2}(x)tan(x)+\frac{m-2}{m-1}\int{sec^{m-2}(x)}dx

    Also, you may want it in terms of x. So, use the fact that {\theta}=sec^{-1}(\frac{x}{4}) and make the sub.
    Last edited by galactus; August 26th 2007 at 08:01 AM.
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