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Thread: 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:

    $\displaystyle
    \int \sqrt{x^2 - 4} dx
    $

    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; Apr 13th 2008 at 03:04 AM.
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  2. #2
    Moo
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    Hello,

    $\displaystyle cosh^2(x)-sinh^2(x)=1$

    $\displaystyle (2cosh(x))^2-(2sinh(x))^2=4$

    $\displaystyle (2cosh(x))^2-4=(2sinh(x))^2$

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

    So $\displaystyle u=2cosh(x)$ 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:

    $\displaystyle
    \int \sqrt{x^2 - 4} dx
    $

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

    Then the integral becomes

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


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


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

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

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

    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,

    $\displaystyle cosh^2(x)-sinh^2(x)=1$

    $\displaystyle (2cosh(x))^2-(2sinh(x))^2=4$

    $\displaystyle (2cosh(x))^2-4=(2sinh(x))^2$

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

    So $\displaystyle u=2cosh(x)$ 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

    $\displaystyle x = 2cosh(t)
    $
    $\displaystyle
    dx/d(t) = 2sinh(t)
    $
    $\displaystyle
    dx = 2sinh(t) d(t)
    $
    $\displaystyle
    \int \sqrt{x^2 - 4} dx
    $
    $\displaystyle
    = \int 2sinh(t)(2sinh(t)) d(t)
    $
    $\displaystyle
    = 2\int 2sinh^2(t) d(t)
    $

    $\displaystyle
    cosh(2t) = 1 + 2sinh^2(t)
    $
    $\displaystyle
    2sinh^2(t) = cosh(2t) - 1
    $
    $\displaystyle
    = 2\int 2sinh^2(t) = 2 \int cosh (2t) - 1 d(t)
    $
    $\displaystyle
    = 2 [sinh(t) - t]
    $

    something went wrong..:S
    Last edited by tasukete; Apr 13th 2008 at 10:23 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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