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Math Help - Integration of a tricky formula

  1. #1
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    Question Integration of a tricky formula

    Can anyone help me solve this?

    Integral of √(4+x^2) / x^5 dx? Thanks.
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  2. #2
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    Quote Originally Posted by nbluo View Post
    Can anyone help me solve this?

    Integral of √(4+x^2) / x^5 dx? Thanks.
    This one's tricky. It requires both a trigonometric substitution and a u substitution.

    First, use the substitution x = 2\tan{\theta}, which means dx = 2\sec^2{\theta}\,d\theta.


    Putting this into the integral...

    \int{\frac{\sqrt{4 + x^2}}{x^5}\,dx} = \int{\frac{\sqrt{4 + (2\tan{\theta})^2}}{(2\tan{\theta})^5}\cdot 2\sec^2{\theta}\,d\theta}

     = \int{\frac{2\sec^2{\theta}\sqrt{4(1 + \tan^2{\theta})}}{32\tan^5{\theta}}\,d\theta}

     = \int{\frac{2\sec^2{\theta}\sqrt{4\sec^2{\theta}}}{  32\tan^5{\theta}}\,d\theta}

     = \int{\frac{4\sec^4{\theta}}{32\tan^5{\theta}}\,d\t  heta}

     = \frac{1}{8}\int{\frac{\sec^4{\theta}}{\tan^5{\thet  a}}\,d\theta}

     = \frac{1}{8}\int{\frac{\cos{\theta}}{\sin^5{\theta}  }\,d\theta}.


    Now we can use a u substitution.

    Let u = \sin{\theta} so that du = \cos{\theta}\,d\theta.

    So the integral becomes

    \frac{1}{8}\int{u^{-5}\,du}

     = -\frac{1}{32}u^{-4} + C

     = -\frac{1}{32}\sin^{-4}{\theta} + C


    Now, since 1 + \cot^2{\theta} = \csc^2{\theta}, we have

    1 + \frac{1}{\tan^2{\theta}} = \sin^{-2}{\theta}

    \left(1 + \frac{1}{\tan^2{\theta}}\right)^2 = \sin^{-4}{\theta}.


    Also remember that x = 2\tan{\theta}, so \tan{\theta} = \frac{x}{2}.


    Thus \sin^{-4}{\theta} = \left(1 + \frac{4}{x^2}\right)^2.


    So finally, the answer to the original integral is...

    -\frac{1}{32}\left(1 + \frac{4}{x^2}\right)^2 + C.
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  3. #3
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    Sorry Prove it, I think you don't solve this pro yet, since when you calculate the square root of 4 sec^2(theta), you got 2 sec(theta). Thanks either.


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  4. #4
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    Quote Originally Posted by nbluo View Post
    Sorry Prove it, I think you don't solve this pro yet, since when you calculate the square root of 4 sec^2(theta), you got 2 sec(theta). Thanks either.


    Ah yes, damn.

    Will edit post if I can solve it...
    Last edited by Prove It; December 9th 2009 at 03:41 PM.
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  5. #5
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    Quote Originally Posted by nbluo View Post
    Can anyone help me solve this?

    Integral of √(4+x^2) / x^5 dx? Thanks.
    put x=\frac1t and your integral becomes -\int{t^{2}\sqrt{4t^{2}+1}\,dt}, now integrate by parts.
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