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Trig substitution problem

  1. Feb 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following integrals or state that they diverge. Use proper notation.

    Integral from 0 to 2 of (x+1)/Square root(4-x^2)

    2. Relevant equations

    3. The attempt at a solution

    I just substituted x = 2sin(theta) thus dx = 2cos(theta)

    I got to the point where it is .5(integral from 0 to 2 of 2tan(theta)) + .5(integral from 0 to 2 of sec(theta)).

    I think theres an easier approach at the problem.

  2. jcsd
  3. Feb 23, 2007 #2
    Partial fractions.
  4. Feb 23, 2007 #3
    I think trig subst. is a good way. The limits of integration will change since you went from x to theta.
  5. Feb 23, 2007 #4


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    So instead of going from 0 to 2, they go from 0 to [itex] \pi/2 [/itex].
  6. Feb 23, 2007 #5
    Ya, that's it. Why do people have to take all the fun out of a problem? Why not leave a hint as a hint? This one-up business is boring.
  7. Feb 23, 2007 #6

    Gib Z

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    [tex]\int\frac{x+1}{4-x^2} dx =\frac{3}{4} \int \frac{1}{2-x} dx - \frac{1}{4}\int \frac{1}{2+x} dx[/tex]. Far too simple from there, the only reason I posted even this far was because I wanted to do it as well :)
  8. Feb 23, 2007 #7

    Gib Z

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    Shoot me in the face, i didnt see the sqrt...
  9. Apr 20, 2007 #8
    hello friend

    ok i will help you as i can


    Integral from 0 to 2 of (x+1)/Square root(4-x^2) dx

    let x=2sin(&)

    =Integral from 0 to [PLAIN]https://www.physicsforums.com/latex_images/12/1253579-0.png [Broken] of (2sin&+1)(2cos&)/2cos& d&

    =Integral from 0 to [PLAIN]https://www.physicsforums.com/latex_images/12/1253579-0.png [Broken] of (2sin&+1) d&

    i think know it is easy to solve

    and i hope that i could help you
    Last edited by a moderator: May 2, 2017
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