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Question using trig substitutions

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate the definite integral


    2. Relevant equations

    3. The attempt at a solution

    I started off this question using the completing the square method, and was able to simplify my integral to the point 17∫1/√(36-u^2) du where u is 5x+2 and du is 5dx. I am fairly certain i am correct up to this point as i have verified this on wolfram alpha. I next subbed in u=6sinv and du=6cosvdv.

    doing this gives me

    (17/36)∫1/(cosv√cos^2(v)) which is just equal to 17/36 ∫ cos^-2(v) which can be written as 17/36 ∫ sec^2(v) and finally gives a result of 17/36 (tanv)

    Knowing that v = arcsin((5x-2)/6) the final answer i got was

    (17/36)(tan(arcsin((5x-2)/6))) which is incorrect.

    I checked with wolfram alpha and through a different method they got 17arcsin((5x+2)/6)

    So basically i have an extra tanv/36 from somewhere, but i cant see where i am making a mistake. can anyone please explain what i may be missing?

    thanks alot
  2. jcsd
  3. Feb 8, 2013 #2


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    Not 36, it was inside a sqrt. And how did that extra cos v get below the line?
  4. Feb 8, 2013 #3
    i replaced du with 6cosv dv so i took the 6 out of the integral and left the extra cosv at the bottom. Why can i not do that?
  5. Feb 8, 2013 #4
    ##\displaystyle 17 \int \frac{1}{\sqrt{36 - u^2}} \ du##

    Subbing u = 6sinv, du = 6cos v dv, we have

    ##\displaystyle 17 \int \frac{1}{\sqrt{36 - 36sin^2v}} 6cosv \ dv = 17 \int \frac{6cosv}{\sqrt{36} \sqrt{1 - sin^2v}} \ dv = 17 \int \frac{cosv}{cosv} \ dv ##

    By the way, you would call this an indefinite integral.
  6. Feb 8, 2013 #5
    sorry the integral you posted didn't show up correctly, do you mind fixing it please? also why is this not an indefinite integral?

    edit: nvm it showed up now, i see what i did wrong, for some reason i tried to replace du with dv/6cosv.

    Thanks alot everybody for the help
  7. Feb 8, 2013 #6
    Sorry about that. I should really test these on TexMaker before I post them. :)
  8. Feb 8, 2013 #7


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    Comment #1: It's rather hard to read through your post with the spacing (or lack thereof) you use, especially since you don't use LaTeX for your integral expressions.

    Comment #2: If I were grading your calculus assignments, you would be losing points right & left for not including the differential . (I'm pretty sure some other Homework Helpers would agree with me, particularly Mark44 .) This omission apparently is what leads to your error below.

    The du should be in your numerator. In the following integral, you put it in the denominator -- well, you put the 6cos(v) in the denominator.

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