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Thread: Integration

  1. #1
    Feb 2009


    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
    Super Member
    Jun 2008
    If you're going to use a trig substitution, then subbing $\displaystyle x = \frac{3}{4}\sec{\theta} \implies dx = \frac{3}{4}\sec{\theta}\tan{\theta} ~d\theta$ would fit.

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

    A neater approach would be to sub $\displaystyle u = \frac{1}{x}$
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  3. #3
    Member javax's Avatar
    Jan 2008
    Milky Way
    Quote Originally Posted by MikeDube View Post
    I start with the integral $\displaystyle dx/[x^2*(sqrt{16x^2-9})]$
    I used trig sub to get $\displaystyle x=3/4 sec(x), dx= sec(x)*tan(x)$ and ended up with the equation...

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

    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; Feb 12th 2009 at 07:10 AM. Reason: additional info
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Mar 2007
    Santiago, Chile
    You can also forget trigo. by putting $\displaystyle x=\frac1u,$ then you'll get an easy integral to solve.
    Last edited by ThePerfectHacker; Feb 12th 2009 at 09:02 AM.
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