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Math Help - integration with substitution of trig/hyperbolic function

  1. #1
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    Exclamation integration with substitution of trig/hyperbolic function

    i was asked to evaluate:

    <br />
\int \sqrt{x^2 - 4} dx<br />

    using a substitution method involving hyperbolic or trig function.

    ive seen some people use x = 2 sec(theta) .ive tried to integrate it but got stuck mid way.

    pls help
    Last edited by tasukete; April 13th 2008 at 04:04 AM.
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  2. #2
    Moo
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    Hello,

    cosh^2(x)-sinh^2(x)=1

    (2cosh(x))^2-(2sinh(x))^2=4

    (2cosh(x))^2-4=(2sinh(x))^2

    \sqrt{(2cosh(x))^2-4}=2sinh(x)

    So u=2cosh(x) seems to be a pretty good idea, plus, its antiderivative is known (sinh(x))
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  3. #3
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    Quote Originally Posted by tasukete View Post
    i was asked to evaluate:

    <br />
\int \sqrt{x^2 - 4} dx<br />

    using a substitution method involving hyperbolic or trig function.

    ive seen some people use x = 2 sec(theta) .ive tried to integrate it but got stuck mid way.

    pls help
    x = 2 \sec t \Rightarrow \frac{dx}{dt} =  -2 \frac{\sin t}{\cos^2 t} \Rightarrow dx = -2 \tan t \, \sec t \, dt.

    Then the integral becomes

    -4 \int \tan t (\tan t \, \sec t) \, dt = -4 \int \tan^2 t \, \sec t \, dt


     = \int (\sec^2 t - 1) \, \sec t \, dt = \int \sec^3 t \, dt - \int \sec t \, dt.


    The second integral can be found in several ways. The result is \ln |\sec t + \tan t|.

    The first requires integration by parts and then the above result.

    Then you have to substitute back that x = 2 \sec t.

    The hyperbolic substitution is easier.
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  4. #4
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    something went wrong

    Quote Originally Posted by Moo View Post
    Hello,

    cosh^2(x)-sinh^2(x)=1

    (2cosh(x))^2-(2sinh(x))^2=4

    (2cosh(x))^2-4=(2sinh(x))^2

    \sqrt{(2cosh(x))^2-4}=2sinh(x)

    So u=2cosh(x) seems to be a pretty good idea, plus, its antiderivative is known (sinh(x))

    i dont know where i went wrong... somehow ended up with -ve intergral.. my final answer came to be -2sinh(theta) + 2theta
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  5. #5
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    Lightbulb

     x = 2cosh(t)<br />
    <br />
dx/d(t) = 2sinh(t)<br />
    <br />
dx = 2sinh(t) d(t)<br />
    <br />
\int \sqrt{x^2 - 4} dx<br />
    <br />
= \int 2sinh(t)(2sinh(t)) d(t)<br />
    <br />
= 2\int 2sinh^2(t) d(t)<br />

     <br />
cosh(2t) = 1 + 2sinh^2(t)<br />
    <br />
2sinh^2(t) = cosh(2t) - 1<br />
    <br />
= 2\int 2sinh^2(t) = 2 \int cosh (2t) - 1  d(t)<br />
    <br />
= 2 [sinh(t) - t]<br />

    something went wrong..:S
    Last edited by tasukete; April 13th 2008 at 11:23 AM.
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  6. #6
    Moo
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    The derivative for x is sinh(t), not -sinh(t)
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  7. #7
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    Cool

    Quote Originally Posted by Moo View Post
    The derivative for x is sinh(t), not -sinh(t)
    oh my..i need sleep thnx heaps ^^
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