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Math Help - Problem with exponential integral

  1. #1
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    Problem with exponential integral



    I'm lost as to how to go about solving this. u = 4x and du = 4 dx but beyond that I'm stumped. That constant in the denominator confuses me.
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    Quote Originally Posted by Archduke01 View Post



    I'm lost as to how to go about solving this. u = 4x and du = 4 dx but beyond that I'm stumped. That constant in the denominator confuses me.

    Hint e^{4x}=\left( e^{2x}\right)^2 Then set u=e^{2x}
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    Quote Originally Posted by TheEmptySet View Post
    Hint e^{4x}=\left( e^{2x}\right)^2 Then set u=e^{2x}
    What would du be? 2?
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    Quote Originally Posted by Archduke01 View Post
    What would du be? 2?
    What is the derivative of e^{\lambda x}?

    if u=e^{\lambda x} \implies \frac{du}{dx}=\lambda \cdot e^{\lambda x}

    Then in terms of differentials you get du=...
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    Quote Originally Posted by TheEmptySet View Post
    What is the derivative of e^{\lambda x}?

    if u=e^{\lambda x} \implies \frac{du}{dx}=\lambda \cdot e^{\lambda x}

    Then in terms of differentials you get du=...
    2e^{2x}!

    1/2 \int 2e^{2x}/(e^{4x} + 64)

    1/2 \int du/(u^2+64)

    ... How can I proceed though?
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    Quote Originally Posted by Archduke01 View Post
    2e^{2x}!

    1/2 \int 2e^{2x}/(e^{4x} + 64)

    1/2 \int du/(u^2+64)

    ... How can I proceed though?
    Now use a trigonometric substitution u = 8\tan{\theta}. Make note that du = 8\sec^2{\theta}\,d\theta.

    Also note that \theta = \arctan{\frac{u}{8}}.


    \int{\frac{1}{u^2 + 64}\,du} = \int{\frac{1}{(8\tan{\theta})^2 + 64}\,8\sec^2{\theta}\,d\theta}

     = \int{\frac{8\sec^2{\theta}}{64\tan^2{\theta} + 64}\,d\theta}

     = \int{\frac{8\sec^2{\theta}}{64(\tan^2{\theta} + 1)}\,d\theta}

     = \int{\frac{\sec^2{\theta}}{8\sec^2{\theta}}\,d\the  ta}

     = \int{\frac{1}{8}\,d\theta}

     = \frac{1}{8}\theta + C

     = \frac{1}{8}\arctan{\frac{u}{8}} + C

     = \frac{1}{8}\arctan{\frac{e^{2x}}{8}} + C.
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    Quote Originally Posted by Prove It View Post
    Now use a trigonometric substitution u = 8\tan{\theta}. Make note that du = 8\sec^2{\theta}\,d\theta.

    Also note that \theta = \arctan{\frac{u}{8}}.
    Thank you for the detailed solution. But why did you choose 8?
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    Quote Originally Posted by Archduke01 View Post
    Thank you for the detailed solution. But why did you choose 8?
    Because if you have an integral of the form

    \int{\frac{1}{a^2 + x^2}\,dx}

    you make the substitution x = a\tan{\theta}.

    This is so that you can factor out the a^2 and then make use of the identity 1 + \tan^2{\theta} = \sec^2{\theta}.

    And since the derivative of \tan{\theta} is \sec^2{\theta}, this also means that the \sec^2{\theta} will be eliminated (as you will end up with them on the top and bottom).
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