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Math Help - Integral of e^sec^2x

  1. #1
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    Integral of e^sec^2x

    Hello,

    I need help evaluating this integral, which i cannot seem to get. It is part of a bigger question(ordinary differential equations)

    The integral is

     \int_(e^{sec^2(x)})
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mkelly09 View Post
    Hello,

    I need help evaluating this integral, which i cannot seem to get. It is part of a bigger question(ordinary differential equations)

    The integral is

     \int_(e^{sec^2(x)})
    Let me start by saying first that - I don't know. Because this feels like e^{u^2} which has no known anti-derivative. And if we were to consider parts, it doesnt look good.

    What about The taylor series for e^x. We could replace x with sec^2x?


    I don't know. Just some thoughts.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mkelly09 View Post
    Hello,

    I need help evaluating this integral, which i cannot seem to get. It is part of a bigger question(ordinary differential equations)

    The integral is

     \int_(e^{sec^2(x)})
    post the original problem in its entirety
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  4. #4
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    OK, original problem is an ordinary differential equation.

    [tex] y' + ytanX = y^2 [/mATH]

    So this is a bernouilli differential equation with a=2

    I set  u(x) = y^{1-a} = 1/y

    So it becomes linear ordinary differential equation by re-writing in terms of u like so:

     u' - (tanx)u = -1

    This is now a linear ODE with  f(x) = -tanx and  r(x) = -1

    Well i don't know this Latex code very well but im going to give it a try:

    The general solution for a linear ODE: <br />
u(x) = \dfrac{-\int {e^{\int{f(x)dx}} * r(x)dx} + c}{e^{\int{f(x)dx}}}<br />

    ill try to figure out this latex code but i gotta run for now, i hope you can all see how this leads me too my original problem posted above.

    <br /> <br />
\int e^{-sec^2(x)}<br /> <br />
    Last edited by mkelly09; January 24th 2010 at 02:12 PM.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mkelly09 View Post
    OK, original problem is an ordinary differential equation.

     y' + ytanX = y^2

    So this is a bernouilli differential equation with a=2

    I set  u(x) = y^{1-a} = 1/y

    So it becomes linear ordinary differential equation by re-writing in terms of u like so:

     u' - (tanx)u = -1

    This is now a linear ODE with  f(x) = -tanx and  r(x) = -1

    Well i don't know this Latex code very well but im going to give it a try:

    The general solution for a linear ODE: <br />
u(x) = \int {(e^{\int {f(x)dx})(r(x))dx} + c<br />

    ill try to figure out this latex code but i gotta run for now, i hope you can all see how this leads me too my original problem posted above.

    <br /> <br />
\int e^{-sec^2(x)}<br /> <br />
    ok, so there's your mistake. i assume you thought \int \tan x ~dx = \sec^2 x + C. This is false. \int \tan x ~dx = \ln | \sec x | + C. what you had before was the derivative.

    so, your integrating factor is \exp \left( - \int \tan x ~dx \right) = \exp ( - \ln |\sec x|) = \cos x

    so your differential equation becomes: \cos x \cdot u' - \sin x \cdot u = - \cos x

    now you should be able to finish up. see how posting the original problem helps?
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    ok, so there's your mistake. i assume you thought \int \tan x ~dx = \sec^2 x + C. This is false. \int \tan x ~dx = \ln | \sec x | + C. what you had before was the derivative.

    so, your integrating factor is \exp \left( - \int \tan x ~dx \right) = \exp ( - \ln |\sec x|) = \cos x

    so your differential equation becomes: \cos x \cdot u' - \sin x \cdot u = - \cos x

    now you should be able to finish up. see how posting the original problem helps?

    Great, thanks so much for your help. This of course did the trick!
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