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Math Help - trig substitution

  1. #1
    Newbie
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    Oct 2009
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    Talking trig substitution

    the question was:
    integral of squareroot(x^2-1) over x^2 dx
    I substituted x for secx(t), and now I have tan^2(t)/sec(t) dt. I have no idea what to do from here, please help!
    EDIT: Now I figured out it can be simplified to the integral of sin^2t/cost dt , or (1-cos(2t))/2cost dt. Still can't figure it out or know which one to continue with...


    Oh, also tried to do another problem,
    integral of x^3 times squareroot(9+4x^2)dx
    I think I'm supposed to substitute 3tan(t) in for x, but then I have the integral of 27tan^3(t) times 3 the squareroot of (1+4tan^2(t)) dt
    and with the 4 in there I can't change it to sec^2(t), so what do I do?

    Any help with either would be much appreciated, thank you!!
    Last edited by ilovecalculus; October 20th 2009 at 07:13 PM. Reason: figured out more
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  2. #2
    Junior Member
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    Doing some rearrangement:

    \int \frac{\sqrt{x^2-1}}{x^2}dx = \int (\frac{1}{\sqrt{x^2-1}} - \frac{1}{x^2 \sqrt{x^2-1}})dx

    The first part is: \int \frac{1}{\sqrt{x^2-1}}dx = cosh^{-1}x + C

    but I tried integration by parts and all sorts of substitutions but still couldn't get that second part.

    However using a calculator I got the answer for the second part and worked back, but this process would not be obvious unless you knew the answer:

    \int \frac{1}{x^2 \sqrt{x^2-1}}dx = \int (\frac{1}{\sqrt{x^2-1}} - \frac{\sqrt{x^2-1}}{x^2})dx  = \int \frac{x^2 - (x^2-1)}{x^2 \sqrt{x^2-1}}dx = \int \frac{\frac{x^2}{\sqrt{x^2-1}}-\sqrt{x^2-1}}{x^2}dx



     = \int \frac{x\frac{d}{dx}(\sqrt{x^2-1})-\frac{d}{dx}(x)\sqrt{x^2-1}}{x^2}dx = \int \frac{d}{dx}(\frac{\sqrt{x^2-1}}{x})dx  = \frac{\sqrt{x^2-1}}{x} + c



    Hence the answer is: \int = cosh^{-1}x - \frac{\sqrt{x^2-1}}{x} + C
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    as for the second part, put x=\frac1t and you'll get a very easy integral to solve.
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