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Thread: weird integral

  1. #1
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    weird integral

    so i have 2pi ∫ (a=1 b= infiniti) 1/x sqrt of 1+ (lnx)^2

    how do i start this beast? trig sub was recommended to me, but i cant see how to start that
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  2. #2
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    I think u sub with u being the ln(x) since du would equal \frac{1}{x} and then trig sub where u=tan
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  3. #3
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    but can you still use trig sub on? ∫du sqrt 1+u^2
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  4. #4
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    Yes, \displaystyle 1+tan^2(\theta)=sec^2(\theta)
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  5. #5
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    oh no im sorry i mean triangle trig, where its like du/sqrt 1+u^2 and u make a triangle with the sqrt on the hyp, u on the tall side and 1 on the base. then find that sec theta = sqrt 1+u^2 tan theta = u sec ^2 theta dtheta = du
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  6. #6
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    Since we know \displaystyle 1+tan^2(\theta)=sec^2(\theta), let \displaystyle u=tan(\theta) \ \mbox{then} \ du=sec^2(\theta)d\theta

    Now by subing, we obtain:

    \displaystyle \int\frac{sec^2(\theta)d\theta}{\sqrt{1+tan^2(\the  ta)}}

    =\displaystyle \int\frac{sec^2(\theta)d\theta}{\sqrt{sec^2(\theta  )}}

    =\displaystyle \int\frac{sec^2(\theta)d\theta}{sec(\theta)}

    =\displaystyle \int sec(\theta)d\theta

    Then integrate sec and remember \displaystyle\theta=tan^{-1}(u) \ \mbox{and} \ u=ln(x)
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  7. #7
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    how did you just let tan theta = u?
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  8. #8
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    We have 1+u^2 and 1+tan^2=sec^2. By making that substitution, we have a well known trig identity. When we make u=tan and substitute, we have 1+(tan)^2=1+tan^2=sec^2. Then by taking the square root, we get sec.
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  9. #9
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    i dont understand how you can just say that u, which = ln|x| can all of a sudden = tan theta
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  10. #10
    MHF Contributor harish21's Avatar
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    You'll probably need to open your calculus book/notes and review integration all over again.
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  11. #11
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    When integrating, we can do as many substitutions as it takes just like we can use L'Hopital's Rule more than once too.
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