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Substitution Question

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data
    By making the substituion [tex]t = \sqrt{1-x}[/tex]

    find [tex]\int \frac{1}{2 + \sqrt{1 - x}}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    So [tex] t = (1-x)^\frac{1/2}[/tex]
    [tex]t' = - \frac{1}{2} (1 - x)^{-\frac{1}{2}}[/tex]

    [tex] dx = -2 \sqrt{1-x} dt [/tex]

    [tex] \int \frac{-2 \sqrt{1-x}}{2 + \sqrt{1-x}} dt [/tex]

    [tex] \int \frac{-2 \sqrt{1-x}}{2 + t} dt [/tex]

    But am I anywhere useful? Am I allowed to say

    [tex] \int \frac{-2t}{2 + t} dt [/tex]

    because I've made the substation already? In that case it's a simple [tex] 2 ln|2+ \sqrt{1-x}|[/tex]

    But that is wrong as the answer is a nasty:

    [tex] 4ln|2+ \sqrt{1-x}| - 2 \sqrt{1-x} + c [/tex]

    Thanks
    Thomas
     
    Last edited: Apr 9, 2010
  2. jcsd
  3. Apr 9, 2010 #2

    Gib Z

    User Avatar
    Homework Helper

    The integral you reached is correct, but You integrated incorrectly. Can you see where you went wrong?
     
  4. Apr 9, 2010 #3
    So

    [tex]
    \int \frac{-2t}{2 + t} dt
    [/tex]

    is right? I'm allowed to say [tex]t = \sqrt{1-x}[/tex] when I change the dx to a [tex]t = \sqrt{1-x}dt[/tex].???

    If so I can see why I can't use ln - because the deravative of the bottom is NOT on the top! Should I integrate by parts?

    Thanks
    Thomas
     
  5. Apr 9, 2010 #4

    Gib Z

    User Avatar
    Homework Helper

     
  6. Apr 9, 2010 #5

    Mark44

    Staff: Mentor

    You started with t = sqrt(1 - x) ==> t2 = 1 - x ==> 2tdt = -dx.

    With this substitution, you can change the original integral wholesale to this one:
    [tex] \int \frac{-2t}{2 + t} dt [/tex]

    Do not use integration by parts here. Your integrand is an improper rational function (the degree of the numerator = the degree of the denominator). Divide the numerator by the denominator to get a simple polynomial and a proper rational function, then integrate.

    Don't forget to undo your substitution.

    Also - don't forget the dx, which you omitted at the start. Leaving off the differential term can come back around and bite you in more complicated problems.
     
  7. Apr 10, 2010 #6

    Sorry I'm a little way off the answer. The substitution [tex]t^{2} = 1 - x[/tex]

    so you've implicitly differentiated that to 2tdt/dx = -1 right

    [tex] \int \frac{-2t}{2 + t} dt [/tex]

    I'm probably just being a complete spanner but the numerator is already divided by tge denominator. Are you trying to get something like

    [tex] ax^{2} + bx + \frac{c}{2+t}[/tex]

    Sorry for the stupidity

    Thomas
     
  8. Apr 10, 2010 #7

    Mark44

    Staff: Mentor

    Actually, I just took the differential of each side. If you multiply your equation by dx, you get my equation.
    No, I'm trying to get an integrand that isn't an improper rational expression (an improper rational expression is one where the degree of the numerator is >= the degree of the denominator). For this integrand, the degree of the numerator is 1, and the degree of the denominator is 1. Divide the numerator by the denominator using polynomial long division. You should get -2 + <some number>/(2 + t).
     
  9. Apr 11, 2010 #8
    Right. Gotcha

    -2 + 4(t+2)

    this integrates to
    -2t + 4ln|t+2|

    wap in the subst

    [tex] 4 ln |2 + \sqrt{1-x} | - 2\sqrt{1-x} +c[/tex]

    It's easy!!!
    Thanks!
     
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