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Thread: Trig Substitution

  1. #1
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    Trig Substitution

    $\displaystyle \int \frac {1}{(x^2+a^2)^{3/2}}dx$

    $\displaystyle \int \frac {1}{(\sqrt{(x^2+a^2)})^3}dx$

    $\displaystyle x=a tan\theta$
    $\displaystyle dx=sec^2\theta $

    $\displaystyle \sqrt{(x^2+a^2)}= \sqrt{a^2tan^2\theta+a^2}= asec\theta\ ?$

    $\displaystyle \int \frac {1}{({asec\theta})^3}sec^2\theta\ ?$
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  2. #2
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    Looks good to me.

    We must remember that $\displaystyle \sqrt{\sec^2\theta}=|\sec\theta|$, but since $\displaystyle \theta=\arctan\frac{x}{a}$, $\displaystyle \sec\theta$ is always positive.
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  3. #3
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    Quote Originally Posted by Scott H View Post
    Looks good to me.

    We must remember that $\displaystyle \sqrt{\sec^2\theta}=|\sec\theta|$, but since $\displaystyle \theta=\arctan\frac{x}{a}$, $\displaystyle \sec\theta$ is always positive.
    Ok thanks, I'm not sure what to do next though. u substitution? trig identities?

    $\displaystyle \int \frac {1}{({asec\theta})^3}sec^2\theta=\ $
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  4. #4
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by JJ007 View Post
    Ok thanks, I'm not sure what to do next though. u substitution? trig identities?

    $\displaystyle \int \frac {1}{({asec\theta})^3}sec^2\theta=\ $


    $\displaystyle \frac{1}{a^3} \int \cos{\theta}d\theta $
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  5. #5
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    Quote Originally Posted by 11rdc11 View Post
    $\displaystyle \frac{1}{a^3} \int \cos{\theta}d\theta $
    $\displaystyle \frac{sin{\theta\ d\theta } }{a^3}+c$
    So is this the full answer to the problem or does it involve a triangle at some point?
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  6. #6
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by JJ007 View Post
    $\displaystyle \frac{sin{\theta\ d\theta } }{a^3}+c$
    So is this the full answer to the problem or does it involve a triangle at some point?
    One more step, back sub for your the trig sub

    $\displaystyle \frac{x}{a^3\sqrt{a^2+x^2}}+C$
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