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Math Help - Integration

  1. #1
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    Integration

    I am struggling with the integral dx/(x^2*sqrt(16x^2-9)). I know the bottom can be broken down using trig substitution to get to x^2*tan^2(theta) but then I get stuck. Any hints on where to go next? Or am I looking at this problem in the wrong direction? Thanks.
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  2. #2
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    If you're going to use a trig substitution, then subbing x = \frac{3}{4}\sec{\theta} \implies dx = \frac{3}{4}\sec{\theta}\tan{\theta} ~d\theta would fit.

    \frac{3}{4}  \int \frac{\sec{\theta}\tan{\theta}}{(\frac{3}{4})^2\se  c^2{\theta}\sqrt{16(\frac{9}{16}\sec^2{\theta})-9}}~d\theta

    A neater approach would be to sub u = \frac{1}{x}
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  3. #3
    Member javax's Avatar
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    Quote Originally Posted by MikeDube View Post
    I start with the integral dx/[x^2*(sqrt{16x^2-9})]
    I used trig sub to get x=3/4 sec(x), dx= sec(x)*tan(x) and ended up with the equation...

    integral of sec(x)tan(x)/(9/16)sec^2(x)*3tan(x)

    After that I pulled out the constants and reduced the fraction until it was only the integral of 1/sec(x)... which is sin(x). This seems way too far off to be the answer, where did I go wrong?
    you did good except when you sub. you should x = 3/4 sec(u). and now your variable under integral will be u. So finally you had 16/9 sin(u). Now you have to turn everything back to x. From your first substitution you can find u, which is u = arcsec(4x/3).
    Last edited by javax; February 12th 2009 at 08:10 AM. Reason: additional info
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  4. #4
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    You can also forget trigo. by putting x=\frac1u, then you'll get an easy integral to solve.
    Last edited by ThePerfectHacker; February 12th 2009 at 10:02 AM.
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