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Tough integral

  1. Oct 24, 2010 #1
    I am trying to figure out which substitution to use to get this integral done:

    [tex]\int \frac{du}{\sqrt{u-u^2} \cdot (1+ub)}[/tex]

    When I plug it into Mathematica I get:

    [tex]\sqrt{\frac{4}{b+1}} \cdot \texttt{arctan} \left ( \sqrt{\frac{(b+1)u}{1-u}} \right )[/tex]

    Any ideas about a suitable substitution?
     
  2. jcsd
  3. Oct 24, 2010 #2

    HallsofIvy

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    Science Advisor

    I would complete the square in the square root: [itex]u- u^2= -(u^2- u+ 1/4- 1/4)= 1/4- (u- 1/2)^2[/itex].

    Then let v= u- 1/2 so that u= v+ 1/2, 1+ ub= (1+ (1/2)b)+ bv.

    Now the integral is
    [tex]\int\frac{dv}{\sqrt{1/4- v^2}((1+ (1/2)b)+ bv)}[/tex]
    and [itex]v= (1/2)sin(\theta)[/itex] looks like a plausible substitution.
     
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