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Using this substitution, show that this integral is equal to the RHS

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Using the substitution [itex]x=acos^2(\theta)+bsin^2(\theta)[/itex]

    [itex]\int^{b}_{a} \sqrt{(x-a)(b-x)}dx = \frac{\pi}{8}(b-a)^2[/itex]


    3. The attempt at a solution

    After making the substitutions and doing all the algebra, I have

    [itex]\int^{\frac{\pi}{2}}_{0} sin(\theta)cos(\theta)(b-a)dx[/itex], with [itex]dx=2(b-a)sin(\theta)cos(\theta)d\theta[/itex]

    Now since the above expression appears as the integrand, shouldn't the integral become [itex]\frac{1}{2}\int^{\frac{\pi}{2}}_{0}d\theta[/itex]? I can only get the RHS if I have [itex]2\int^{\frac{\pi}{2}}_{0} sin^2(\theta)cos^2(\theta)(b-a)^2d\theta[/itex]
     
    Last edited: May 27, 2013
  2. jcsd
  3. May 27, 2013 #2

    Ray Vickson

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    Are you sure you have the right problem? Should it be ##\int_a^b \sqrt{(x-a)(b-x)}\, dx?##
     
  4. May 27, 2013 #3
    Oh yeah, typo. Fixed it, thanks :p
     
  5. Jun 4, 2013 #4
    Solutions haven't been provided for this particular problem set for some reason so I'm still curious as to whether the method is correct :p
     
  6. Jun 5, 2013 #5

    Mark44

    Staff: Mentor

    You shouldn't need a solution. They have given you what the left side of the equation is supposed to be equal to, provided you haven't made mistakes in your work.
     
  7. Jun 5, 2013 #6
    I probably should've said I don't understand the method. I don't know why we keep both the substitution and the differential substitution in the integrand
     
  8. Jun 5, 2013 #7

    haruspex

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    You correctly have this:
    [itex]\int^{\frac{\pi}{2}}_{0} sin(\theta)cos(\theta)(b-a)dx[/itex], where [itex]dx=2(b-a)sin(\theta)cos(\theta)d\theta[/itex]
    From there, I don't understand the difficulty. You use the second expression to replace dx in the integral and get exactly what you wanted.
     
  9. Jun 5, 2013 #8
    Ah, I see where I went wrong now... usually in evaluating something like [itex]\int cos(x)(1-sin^2(x))[/itex], we let u=sin(x) and so du=cos(x)dx and since it's already under the integral, we write [itex]\int 1-u^2 du[/itex]. I blindly thought it was the same for this one and didn't know why we didn't get rid of the integrand when making the substitution.
     
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