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Math Help - I need help on this integral

  1. #1
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    Exclamation I need help on this integral

    Never mind. I got the solution.
    Last edited by Locke; February 23rd 2010 at 04:54 PM.
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  2. #2
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    Quote Originally Posted by Locke View Post
    1 / (9 + x^2)^(3/2) dx

    I actually got the answer to this integral here, but I have no idea how to get it (I need to know the steps).

    Any ideas?
    \int{\frac{1}{(9 + x^2)^{\frac{3}{2}}}\,dx}.


    Make the substitution x = 3\sinh{t} so that dx = 3\cosh{t}\,dt.

    It's important to note that from our original substitution: \sinh{t} = \frac{x}{3}.


    So the integral becomes

    \int{\frac{1}{[9 + (3\sinh{t})^2]^{\frac{3}{2}}}\,3\cosh{t}\,dt}

     = \int{\frac{3\cosh{t}}{[9(1 + \sinh^2{t})]^{\frac{3}{2}}}\,dt}

     = \int{\frac{3\cosh{t}}{27(\sqrt{\cosh^2{t}})^3}\,dt  }

     = \frac{1}{9}\int{\frac{\cosh{t}}{\cosh^3{t}}\,dt}

     = \frac{1}{9}\int{\frac{1}{\cosh^2{t}}\,dt}

     = \frac{1}{9}\tanh{t} + C

     = \frac{\sinh{t}}{9\cosh{t}} + C

     = \frac{\sinh{t}}{9\sqrt{1 + \sinh^2{t}}} + C.

     = \frac{\frac{x}{3}}{9\sqrt{1 + \left(\frac{x}{3}\right)^2}} + C

     = \frac{\frac{x}{3}}{9\sqrt{\frac{9 + x^2}{9}}} + C

     = \frac{\frac{x}{3}}{\frac{9\sqrt{9 + x^2}}{3}} + C

     = \frac{\frac{x}{3}}{3\sqrt{9 + x^2}} + C

     = \frac{x}{9\sqrt{9 + x^2}} + C.



    Another alternative is to make the substitution x = 3\tan{\theta}.
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  3. #3
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    What does sinh mean? I did it by substituting x for 3tan and got the same answer.
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  4. #4
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    Quote Originally Posted by Locke View Post
    What does sinh mean? I did it by substituting x for 3tan and got the same answer.
    The Hyperbolic sine.

    Hyperbolic function - Wikipedia, the free encyclopedia


    And like I said, substituting x = 3\tan{\theta} should give you the same answer.
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