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Question regarding u-substitution

  1. Apr 17, 2015 #1
    1. The problem statement, all variables and given/known data
    I guess it's actually trig substitution...
    To demonstrate I'll use the integral of $$ \int \dfrac{x \cdot \ln(x)}{\sqrt{x^2-1}} \cdot dx$$

    2. Relevant equations
    3. The attempt at a solution

    $$ x=\sec\theta \\ dx=\sec\theta \cdot \tan\theta \\ \int \dfrac{\sec\theta \cdot \ln\sec\theta \cdot \sec\theta \cdot \tan\theta}{\sqrt{\sec^2\theta-1}} \cdot d\theta$$
    $$ \int \sec^2\theta \cdot \ln\sec\theta \cdot d\theta $$
    Everything in this equation is in terms of ## \sec\theta ## which is what I said x is equal to earlier. However, if I sub x back in for ##\sec\theta## and do integration by parts I end up getting the wrong answer. In the answer key they do the integration by parts with the ##\theta## variable and when they change back to x get a completely different answer than me.
    For reference the answer I got was ## \dfrac{x^3 \cdot lnx}{3} - \dfrac{x^3}{9} + C ## and their answer is ## ln|x| \sqrt{x^2-1} - \sqrt{x^2-1} + arcsec(x)+ C ##
    Okay so I think my mistake is that I did not replace ## d\theta ## in terms of ##dx##
     
    Last edited: Apr 17, 2015
  2. jcsd
  3. Apr 17, 2015 #2

    SteamKing

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    If you are going to use integration by parts anyway, taking a detour thru the Trig Zone :eek: may not be the ideal approach for this integral.

    Have you tried taking the integrand as is and finding a suitable u and dv for IBP?
     
  4. Apr 17, 2015 #3

    SammyS

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    Yes, forgetting to handle dθ with care would cause a problem.

    As you might expect, that will simply change the integral back to its original form.
     
  5. Apr 17, 2015 #4
    Yea although I personally find it easier to integrate after using substitutions to get rid of the roots.
     
  6. Apr 17, 2015 #5

    SteamKing

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    Looks like there's roots in the book's solution. :rolleyes:

    In any event, you should always differentiate what you obtain for the integral (or even what the book says) to check that it returns the original integrand. :wink:
     
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