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Math Help - Integration help: 7*sqrt(x^2+9)

  1. #1
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    Integration help: 7*sqrt(x^2+9)

    I need to integrate the following:

    7*sqrt(x^2+9)

    Here is my work so far:

    x=3tan(theta)
    d-x=3sec^2(theta)*d-theta

    Then:
    7*int[sqrt{9tan^2(theta)+9}*3sec^2(theta)*d-theta

    Then:
    63*int[sec^3(theta)*d-theta]
    And I don't know what to do from there. Thank you!
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  2. #2
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    I'm pretty sure you can solve the integral of sec^3 by doing integration by parts

    Take sec^2 and sec as your two parts and try that
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  3. #3
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    stupid!!! Ah I feel dumb. I need sleep :P

    Thanks man!
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  4. #4
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    Quote Originally Posted by dillonmhudson View Post
    I need to integrate the following:

    7*sqrt(x^2+9)

    Here is my work so far:

    x=3tan(theta)
    d-x=3sec^2(theta)*d-theta

    Then:
    7*int[sqrt{9tan^2(theta)+9}*3sec^2(theta)*d-theta

    Then:
    63*int[sec^3(theta)*d-theta]
    And I don't know what to do from there. Thank you!
    \int{7\sqrt{x^2 + 9}\,dx}

    Use a hyperbolic substitution instead - it's easier...

    Substitute x = 3\sinh{t} so that dx = 3\cosh{t}\,dt.

    Note that t = \sinh^{-1}{\frac{x}{3}}.


    The integral becomes

    \int{7\sqrt{(3\sinh{t})^2 + 9}\cdot 3\cosh{t}\,dt}

     = \int{21\cosh{t}\sqrt{9\sinh^2{t} + 9}\,dt}

     = \int{21\cosh{t}\sqrt{9(\sinh^2{t} + 1)}\,dt}

     = \int{63\cosh{t}\sqrt{\cosh^2{t}}\,dt}

     = 63\int{\cosh^2{t}\,dt}

     = \frac{63}{2}\int{\cosh{2t} + 1\,dt}

     = \frac{63}{2}\left[\frac{1}{2}\sinh{2t} + t\right] + C

     = \frac{63}{2}\left[\sinh{t}\cosh{t} + t\right] + C

     = \frac{63}{2}\left[\sinh{t}\sqrt{1 + \sinh^2{t}} + t\right] + C

     = \frac{63}{2}\left[\frac{x}{3}\sqrt{1 + \left(\frac{x}{3}\right)^2} + \sinh^{-1}{\frac{x}{3}}\right] + C

     = \frac{7x\sqrt{9 + x^2}}{2} + \frac{63\sinh^{-1}{\frac{x}{3}}}{2} + C
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  5. #5
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    how about a solution without hyperbolics?
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