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Math Help - Integral of sqrt(1 + e^(2x))

  1. #1
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    Integral of sqrt(1 + e^(2x))

    Integral of sqrt(1 + e^(2x))

    u = 1 + e^(2x) ??

    Please point me in the right direction. Thank you.
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  2. #2
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    Quote Originally Posted by jsel21 View Post
    Integral of sqrt(1 + e^(2x))

    u = 1 + e^(2x) ??

    Please point me in the right direction. Thank you.
    1+\tan^2=\sec^2

    e^{2x}=\left(e^x\right)^2

    e^x=\tan\theta

    Try this.
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  3. #3
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    Quote Originally Posted by jsel21 View Post
    Integral of sqrt(1 + e^(2x))

    u = 1 + e^(2x) ??

    Please point me in the right direction. Thank you.
    integrate Sqrt[1 + Exp[2x]] - Wolfram|Alpha+

    Click on Show steps.
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    \displaystyle \begin{aligned} \int \sqrt{1+e^{2x}}\;{dx} & = \int\frac{\sqrt{1+e^{2x}}\sqrt{1+e^{2x}}}{\sqrt{1+  e^{2x}}}\;{dx} \\&= \int\frac{1+e^{2x}}{\sqrt{1+e^{2x}}}\;{dx} \\& = \int\frac{1}{\sqrt{1+e^{2x}}}\;{dx}+\int\frac{e^{2  x}}{\sqrt{1+e^{2x}}}\;{dx} <br />
\\& = I_{1}+\int \frac{(1+e^{2x})'}{2\sqrt{1+e^{2x}}}\;{dx} \\& = I_{1}+\sqrt{1+e^{2x}}+k.\end{aligned}

    Where

    \displaystyle I_{1} = \int\frac{1}{\sqrt{1+e^{2x}}}\;{dx}.

    Let t = \sqrt{1+e^{2x}}, so that dx = \frac{\sqrt{1+e^{2x}}}{e^{2x}}\;{dt} = \frac{t}{t^2-1}\;{dt}, then:

    \displaystyle \begin{aligned} I_{1} & = \int\frac{t}{t(t^2-1)}\;{dt} \\& =  \int\frac{1}{(t-1)(t+1)}\;{dt} \\ & = \frac{1}{2}\int \frac{(t+1)-(t-1)}{(t+1)(t-1)}\;{dt} \\& = \frac{1}{2}\int\frac{1}{t-1}\;{dt}-\frac{1}{2}\int\frac{1}{t+1}\;{dt} \\& = \frac{1}{2}\ln(t-1)-\frac{1}{2}\ln(t+1)+k \\ & = \frac{1}{2}\ln\left\{\frac{t-1}{t+1}\right\}+k \\ & = \frac{1}{2}\ln\left\{\frac{\sqrt{1+e^{2x}}-1}{\sqrt{1+e^{2x}}+1}\right\}+k_{1}. \end{aligned}

    Thus \displaystyle \boxed{\int \sqrt{1+e^{2x}}\;{dx} =  \frac{1}{2}\ln\left\{\frac{\sqrt{1+e^{2x}}-1}{\sqrt{1+e^{2x}}+1}\right\}+\sqrt{1+e^{2x}}+k}.
    Last edited by TheCoffeeMachine; February 8th 2011 at 05:18 PM.
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    1+\tan^2=\sec^2

    e^{2x}=\left(e^x\right)^2

    e^x=\tan\theta

    Try this.
    Interesting...

    \displaystyle \int{\sqrt{1 + e^{2x}}\,dx}

    Make the substitution \displaystyle e^x = \tan{\theta} \implies x = \ln{(\tan{\theta})} \implies dx = \frac{\sec^2{\theta}}{\tan{\theta}}\,d\theta and the integral becomes

    \displaystyle \int{\sqrt{1 + \tan^2{\theta}}\,\frac{\sec^2{\theta}}{\tan{\theta  }}\,d\theta}

    \displaystyle = \int{\sec{\theta}\,\frac{\sec^2{\theta}}{\tan{\the  ta}}\,d\theta}

    \displaystyle = \int{\sec^3{\theta}\cot{\theta}\,d\theta}

    \displaystyle = \int{\frac{1}{\cos^3{\theta}}\,\frac{\cos{\theta}}  {\sin{\theta}}\,d\theta}

    \displaystyle = \int{\frac{1}{\sin{\theta}\cos^2{\theta}}\,d\theta  }

    \displaystyle = \int{\frac{\sin{\theta}}{\sin^2{\theta}\cos^2{\the  ta}}\,d\theta}

    \displaystyle = \int{\frac{\sin{\theta}}{(1 - \cos^2{\theta})\cos{\theta}}\,d\theta}

    \displaystyle = \int{\frac{-\sin{\theta}}{(\cos^2{\theta} - 1)\cos{\theta}}\,d\theta}

    \displaystyle = \int{\frac{-\sin{\theta}}{(\cos{\theta} - 1)(\cos{\theta} + 1)\cos{\theta}}\,d\theta}.

    And now making the substitution \displaystyle u = \cos{\theta} \implies du = -\sin{\theta}\,d\theta the integral becomes

    \displaystyle \int{\frac{1}{(u-1)(u+1)u}\,du}

    which can be solved using partial fractions
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Interesting...

    \displaystyle \int{\sqrt{1 + e^{2x}}\,dx}

    Make the substitution \displaystyle e^x = \tan{\theta} \implies x = \ln{(\tan{\theta})} \implies dx = \frac{\sec^2{\theta}}{\tan{\theta}}\,d\theta and the integral becomes

    \displaystyle \int{\sqrt{1 + \tan^2{\theta}}\,\frac{\sec^2{\theta}}{\tan{\theta  }}\,d\theta}

    \displaystyle = \int{\sec{\theta}\,\frac{\sec^2{\theta}}{\tan{\the  ta}}\,d\theta}

    \displaystyle = \int{\sec^3{\theta}\cot{\theta}\,d\theta}

    \displaystyle = \int{\frac{1}{\cos^3{\theta}}\,\frac{\cos{\theta}}  {\sin{\theta}}\,d\theta}

    \displaystyle = \int{\frac{1}{\sin{\theta}\cos^2{\theta}}\,d\theta  }

    \displaystyle = \int{\frac{\sin{\theta}}{\sin^2{\theta}\cos^2{\the  ta}}\,d\theta}
    That was the first thing that came to mind.
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    Thanks guys, these posts were very helpful.
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    \displaystyle\int{\sqrt{1+{{e}^{2x}}}\,dx}=\int{\f  rac{\sqrt{1+{{e}^{2x}}}{{e}^{2x}}}{{{e}^{2x}}}\,dx  }=\int{\frac{{{t}^{2}}}{{{t}^{2}}-1}\,dt}, there's no much here, just clearing out the clutter and it's easy.
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  9. #9
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    Also

    \displaystyle\int\frac{dt}{\sin t\cos^2 t}=\int\frac{\sin^2 t+\cos^2}{\sin t\cos^2 t}dt=\int\frac{\sin t}{\cos^2 t}dt+\int\csc tdt
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  10. #10
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    take u^2 = 1 + e^(2x)
    so 2udu = 2e^(2x)
    which is 2(u^2-1)dx
    2udu = 2(u^2-1)dx
    dx = u/(u^2-1) du
    put it in the question.. it will be u^2/(u^2-1) du = (u^2 - 1 + 1)/(u^2-1)du
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  11. #11
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    you did the same as me.
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    hmm.. sorry
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  13. #13
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    Quote Originally Posted by Prove It View Post
    \displaystyle \int{\sqrt{1 + \tan^2{\theta}}\,\frac{\sec^2{\theta}}{\tan{\theta  }}\,d\theta}

    \displaystyle = \int{\sec{\theta}\,\frac{\sec^2{\theta}}{\tan{\the  ta}}\,d\theta}
    I hate it when trig substitutions do this kind of thing.
    sec( \theta ) = \pm \sqrt{1 + tan^2 ( \theta )}

    But the original integrand is positive definite, so we can discard the negative solution. Correct? Or is it more complicated than that?

    -Dan
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  14. #14
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    In this case, on Prove's substitution, that forces to 0\le\theta<\dfrac\pi2 and we have that |\sec\theta|=\sec\theta.

    Of course we coulda taken \pi\le\theta<\dfrac32\pi but that makes \sec\theta non positive.

    It all depends on how you fix the angle, but that's more a matter of dealing with definite integrals.
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