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Math Help - Integral- trig substitution?

  1. #1
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    Integral- trig substitution?

    Hi. I'm trying to solve this integral. I think I may need a trig substitution, but I've tried a ton and I'm lost. Any specific help would be much appreciated.

    \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx


    The solution is


    \frac{x}{z^2(x^2+z^2)^{1/2}}





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  2. #2
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    Quote Originally Posted by daberkow View Post
    Hi. I'm trying to solve this integral. I think I may need a trig substitution, but I've tried a ton and I'm lost. Any specific help would be much appreciated.

    \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx

    The solution is

    \frac{x}{z^2(x^2+z^2)^{1/2}}
    Assuming z is a constant:
    Yes. Triginometric Substitution. Let x=ztan(\theta) then dx=zsec^2(\theta)d\theta.
    Post your work; to see why did you stuck.
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  3. #3
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    <br />
      \int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx<br />

    \int\frac{zsec^2\theta}{(z^2+z^2tan^2\theta)^\frac  {3}{2}}d\theta

    \int\frac{zsec^2\theta}{((z^2(1+tan^2\theta))^\fra  c{3}{2}}d\theta

    \int\frac{zsec^2\theta}{(z^2(sec^2\theta))^\frac{3  }{2}}d\theta

    \int\frac{sec^2\theta}{z^2(sec^2\theta))^\frac{3}{  2}}d\theta

    \int\frac{1}{z^2(sec^2\theta)^\frac{1}{2}}d\theta

    \int\frac{1}{z^2(sec\theta)}d\theta

    I went wrong somewhere...
    Last edited by daberkow; February 23rd 2010 at 03:20 PM.
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  4. #4
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    Quote Originally Posted by daberkow View Post
    <br />
\int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx<br />

    \int\frac{zsec^2\theta}{(z^2+z^2tan^2\theta)^\frac  {3}{2}}d\theta

    \int\frac{zsec^2\theta}{((z^2(1+tan^2\theta))^\fra  c{3}{2}}d\theta

    \int\frac{zsec^2\theta}{(z^2(sec^2\theta))^\frac{3  }{2}}d\theta

    \int\frac{sec^2\theta}{z^2(sec^2\theta))^\frac{3}{  2}}d\theta

    \int\frac{1}{z^2(sec^2\theta)^\frac{1}{2}}d\theta

    \int\frac{1}{z^2(sec\theta)}d\theta

    I went wrong somewhere...
    Try using \frac{1}{\sec\theta}=\cos\theta
    Last edited by ione; February 23rd 2010 at 06:10 PM.
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  5. #5
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    Quote Originally Posted by daberkow View Post
    I went wrong somewhere...
    No, keep going...

    http://www.mathhelpforum.com/math-he...tml#post457096
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  6. #6
    Ted
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    Quote Originally Posted by daberkow View Post
    <br />
\int\frac{1}{(x^2+z^2)^\frac{3}{2}}dx<br />

    \int\frac{zsec^2\theta}{(z^2+z^2tan^2\theta)^\frac  {3}{2}}d\theta

    \int\frac{zsec^2\theta}{((z^2(1+tan^2\theta))^\fra  c{3}{2}}d\theta

    \int\frac{zsec^2\theta}{(z^2(sec^2\theta))^\frac{3  }{2}}d\theta

    \int\frac{sec^2\theta}{z^2(sec^2\theta))^\frac{3}{  2}}d\theta

    \int\frac{1}{z^2(sec^2\theta)^\frac{1}{2}}d\theta

    \int\frac{1}{z^2(sec\theta)}d\theta

    I went wrong somewhere...
    Good. No mistakes.
    You almost done.
    \int\frac{1}{z^2(sec\theta)}d\theta=\frac{1}{z^2}\  int cos(\theta) d\theta
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