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Math Help - integration by substitution

  1. #1
    Member disclaimer's Avatar
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    integration by substitution

    Hi, can you think of a way to solve the following integral using substitution?

    \int{e^{-x}\arctan{e^x}}dx

    Help much appreciated.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by disclaimer View Post
    Hi, can you think of a way to solve the following integral using substitution?

    \int{e^{-x}\arctan{e^x}}dx

    Help much appreciated.
    i don't think substitution alone will work.

    we can begin with substitution.

    a substitution of u = e^x yields \int \frac {\arctan u}{u^2}~du

    which we can do by parts
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  3. #3
    Super Member wingless's Avatar
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    Quote Originally Posted by Jhevon View Post
    i don't think substitution alone will work.

    we can begin with substitution.

    a substitution of u = e^x yields \int \frac {\arctan u}{u^2}~du

    which we can do by parts
    How did you use the substitution to get \int \frac {\arctan u}{u^2}~du ?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wingless View Post
    How did you use the substitution to get \int \frac {\arctan u}{u^2}~du ?
    oh, did i make a mistake?

    we have \int \frac {\arctan e^x}{e^x}~dx

    = \int \frac {e^x \arctan e^x}{e^{2x}}~dx

    Let u = e^x

    \Rightarrow du = e^x~dx

    So our integral becomes

    \int \frac {\arctan u}{u^2}~du
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  5. #5
    Super Member wingless's Avatar
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    Ah, OK, my bad.. I saw the integral wrong..
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  6. #6
    Math Engineering Student
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    Quote Originally Posted by disclaimer View Post
    \int{e^{-x}\arctan{e^x}}dx
    Start by integrating by parts:

    \int {e^{ - x} \arctan e^x \,dx}  =  - e^{ - x} \arctan e^x  + \underbrace {\int {\frac{1}<br />
{{e^{2x}  + 1}}\,dx} }_\lambda .
    Now \lambda can be evaluated as follows:

    \lambda  = \int {\frac{{e^{2x}  + 1 - e^{2x} }}<br />
{{e^{2x}  + 1}}\,dx}  = x - \frac{1}<br />
{2}\ln \left( {e^{2x}  + 1} \right) + k .

    The full answer is

    \int {e^{ - x} \arctan e^x \,dx}  =  - e^{ - x} \arctan e^x  + x - \frac{1}<br />
{2}\ln \left( {e^{2x}  + 1} \right) + k.
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