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Integration with trig substitution

  1. Mar 7, 2016 #1
    1. The problem statement, all variables and given/known data
    The problem is the integral attached

    2. Relevant equations
    sec2(u)=(1+tan2(x))
    a2+b2=c2
    ∫cos(u)=-sin(u)+C

    3. The attempt at a solution
    The solution is attached. I am wondering if someone could give me a hint where I went drastically wrong or where I possibly dropped a negative. Hopefully what will come of this is the latter.

    Thank you in advance,

    Ethan
     

    Attached Files:

  2. jcsd
  3. Mar 7, 2016 #2
    The one idea I have, by the way, is when I simplify the √sec2(x), it could become -sec(x). This is my only idea.
     
  4. Mar 7, 2016 #3

    SammyS

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    Yes, if sec(x) < 0, then ##\ \sqrt{\sec^2(x)\,}=-\sec (x)\ .##
     
  5. Mar 7, 2016 #4
    how do I know sec(x)< 0? The only information given is the integral at the beginning and that a>0.
     
  6. Mar 7, 2016 #5

    Mark44

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    I see two mistakes.
    1. ##\int cos u du = sin u##, not -sin u
    2. ##sec u = \frac {\sqrt{a^2 + x^2}}{a}##. You don't have that written down, and I suspect that you lost a factor of a in your simplification.
    No, that would be plain old sec(x).
     
  7. Mar 7, 2016 #6

    Mark44

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    If ##0 \le u < \pi/2##, sec(u) will be positive. This is a reasonable assumption in your trig substitution.
     
  8. Mar 7, 2016 #7
    Thank you,

    My only follow up question is why is this significant?

    I am not at sin(u)/a. I know by the trig circle that the opposite side is equal to x, the adjacent side is a, and the hypotenuse is √(x2+a2 ). This means sin(u)=x/√(x2+a2 ). I don't know where the sec(u) back substitution came from.

    Thank You,

    Ethan
     
  9. Mar 8, 2016 #8

    Mark44

    Staff: Mentor

    I always draw a triangle when I do trig substitution -- I don't clutter up my brain with memorized forumulas that I might forget. In my triangle u is the acute angle I'm interested in. The adjacent side is a, the opposite side is x, the hypotenuse is ##\sqrt{a^2 + x^2}##. From this triangle I get ##tan u = \frac x a## and my expression for sec(u).
     
  10. Mar 8, 2016 #9
    I guess my main question is why are you dealing with sec(u) when the question ends up with sin(u)/a2. Shouldn't I be drawing a trig ciricle for sin(u) and not sec(u)?
     
  11. Mar 8, 2016 #10

    Mark44

    Staff: Mentor

    The sec(u) factor comes in because of the ##(a^2 + x^2)^{3/2}## in the denominator. In the triangle I described, ##\tan u = \frac x a## and ##\sec u = \frac{\sqrt{a^2 + x^2}}{a}##. I hope that's clear.

    You use the triangle (not a trig "ciricle") to undo the substitution ##\tan u = \frac x a##. From the triangle you can figure out all of the trig relationships, including ##\sin u##.
     
  12. Mar 8, 2016 #11

    vela

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    I would suggest omitting some of the unnecessary algebraic manipulations, like what you did with the denominator before doing the trig substitution.
    $$(a^2+x^2)^{3/2} = (\sqrt{a^2+x^2})^3 = (\sqrt{a^2+a^2\tan^2 u})^3 = (a\sec u)^3$$ Fewer steps, easier to read for the grader, less chance of making a dumb algebra mistake, etc. Also, after you change variables to ##u##, the limits are no longer ##0## and ##a##. If you had changed the limits to ##u=0## to ##u=\pi/4##, you probably would have realized that ##\sec u > 0## and that your error lay elsewhere.
     
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