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Thread: 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?

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

    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?

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

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

    we can begin with substitution.

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

    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 $\displaystyle u = e^x$ yields $\displaystyle \int \frac {\arctan u}{u^2}~du$

    which we can do by parts
    How did you use the substitution to get $\displaystyle \int \frac {\arctan u}{u^2}~du$ ?
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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 $\displaystyle \int \frac {\arctan u}{u^2}~du$ ?
    oh, did i make a mistake?

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

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

    Let $\displaystyle u = e^x$

    $\displaystyle \Rightarrow du = e^x~dx$

    So our integral becomes

    $\displaystyle \int \frac {\arctan u}{u^2}~du$
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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
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    Quote Originally Posted by disclaimer View Post
    $\displaystyle \int{e^{-x}\arctan{e^x}}dx$
    Start by integrating by parts:

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

    $\displaystyle \lambda = \int {\frac{{e^{2x} + 1 - e^{2x} }}
    {{e^{2x} + 1}}\,dx} = x - \frac{1}
    {2}\ln \left( {e^{2x} + 1} \right) + k .$

    The full answer is

    $\displaystyle \int {e^{ - x} \arctan e^x \,dx} = - e^{ - x} \arctan e^x + x - \frac{1}
    {2}\ln \left( {e^{2x} + 1} \right) + k.$
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