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Math Help - Trignometric Integrals

  1. #1
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    Trignometric Integrals

    <br /> <br />
\int \sqrt{x^2 - 4} dx<br /> <br />

    <br /> <br />
\int \frac{e^x}{1+e^2x} dx <br /> <br />

    'e' is for Euler's number.

    I'm having a serious problem with these. I just can't get it.
    Can you also show the steps at how you arrived at the answer.
    Thank you so much!
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  2. #2
    Member Nacho's Avatar
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    To the first you can do: \sec \theta=\dfrac{x}{2} and to the second u=e^x
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  3. #3
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    Quote Originally Posted by Nacho View Post
    To the first you can do: \sec \theta=\dfrac{x}{2} and to the second u=e^x
    i ended up with
    <br />
4 \int \frac{sin^2\theta}{cos^3\theta} d\theta<br />

    i dont know how else to go from there
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  4. #4
    Super Member General's Avatar
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    Quote Originally Posted by larryboi7 View Post
    i ended up with
    <br />
4 \int \frac{sin^2\theta}{cos^3\theta} d\theta<br />

    i dont know how else to go from there
    4 \int \frac{sin^2\theta}{cos^3\theta} d\theta = 4\int tan^2(\theta) sec(\theta) d\theta
    = 4 \int (sec^2(\theta)-1)sec(\theta) d\theta = 4 \int sec^3(\theta) d\theta - 4 \int sec(\theta) d\theta = ....

    Can you complete it ?
    Use integration by parts for the first one.

    For the second integral in your question:
    Its e^2x or e^{2x} ?
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  5. #5
    MHF Contributor Calculus26's Avatar
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    the details of the first are given in a previous post of yours "competing the integration"
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  6. #6
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    Quote Originally Posted by Calculus26 View Post
    the details of the first are given in a previous post of yours "competing the integration"
    Normally I'd close one of the threads but in this instance it's too difficult to straighten things out so I'm leaving both threads open.

    Please post replies here that are only related to \int \frac{e^x}{1+e^{2x}} dx. Replies to the other one should be posted here: http://www.mathhelpforum.com/math-help/calculus/128484-completing-integration.html
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