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Homework Help: Integration: completing the square and inverse trig functions

  1. May 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\int[/tex](x+2)dx/sqrt(3+2x-x2)

    2. Relevant equations

    [tex]\int[/tex]du/sqrt(a2 - u2) = sin-1u/a

    3. The attempt at a solution

    I began by completing the square:
    3+2x-x2 = 4 -(x2-2x+1)

    So, 4-(x-1)2 = a2-u2 and a=2 and u=(x-1)
    Further, since x=(u+1), dx=du and (x+2)=(u+3)

    Substituting, I got:

    [tex]\int[/tex](x+2)dx/sqrt(3+2x-x2) = [tex]\int[/tex](u+3)du/sqrt(a2-u2)

    = [tex]\int[/tex]udu/sqrt(a2-u2) + 3[tex]\int[/tex]du/sqrt(a2-u2)

    This is where I get confused. I know that 3[tex]\int[/tex]du/sqrt(a2-u2) = 3sin-1u/a

    But, according to my book, [tex]\int[/tex]udu/sqrt(a2-u2) = -sqrt(a2-u2) and I don't understand why.

    Can someone offer an explanation? Thanks.
  2. jcsd
  3. May 29, 2010 #2

    using substitution

    cos [tex]\theta[/tex] = [tex]\frac{u}{a}[/tex]
  4. May 29, 2010 #3
    Are you saying that I should substitute u=a sin theta to obtain:

    [tex]\int[/tex]udu/sqrt(a2-u2) = [tex]\int[/tex](a sin theta)(a cos theta)(d theta)/(a cos theta) = a cos theta

    So, u = a cos theta and cos theta = u/a, but u/a doesn't equal -sqrt(a2-u2)

    What am I missing?
  5. May 30, 2010 #4

    see the difference :
    3[tex]\int[/tex]du/sqrt(a2-u2) = 3sin-1u/a


    [tex]\int[/tex]u du/sqrt(a2-u2) = -sqrt(a2-u2)

    to obtain the second one, use : v = a2-u2
  6. May 30, 2010 #5


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    Homework Helper

    Just take the derivative of both answers and you'll see why.
  7. May 30, 2010 #6
    I see it now. Thanks for your help.


    Substitute v=a2-u2, dv=-2udu, udu=-dv/2

    = -1/2[tex]\int[/tex]v-1/2dv = -v-1/2 = -sqrt(a2-u2)
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