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Integration help

  1. Apr 12, 2005 #1
    I was doing a change in variables problem in multivariable calculus and I got stuck on the last integration.

    [tex]\frac {4}{3} \int_{u=0}^{1} (1-u^2)^{\frac {1}{2}}(2u^2+1)du[/tex]

    I don't think substitution works. Can anyone show me an easy way to solve this? Thanks.
     
  2. jcsd
  3. Apr 12, 2005 #2

    dextercioby

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    Typically [itex] u=\sin t [/itex]

    Daniel.
     
  4. Apr 13, 2005 #3
    For problems like this in general, you should try spliting up any integrand into as many pieces as possible...

    Here you can rewrite the integrand obviously as 2u^2[(1-u^2)^0.5] + (1-u^2)...

    and being the lazy engineering student that I am... I would look these up in a table of integrals--- which would definetely have the general solutions.
     
  5. Apr 13, 2005 #4

    dextercioby

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    Don't listen to an engineering student giving advice in anything but engineering...:wink:


    The transformed integral should be

    [tex] \frac{4}{3}\int_{0}^{\frac{\pi}{2}} \cos^{2}t\left(2\sin^{2}t+1\right) dt [/tex]

    Use the double angle formulas to get it simplified.

    Daniel.
     
  6. Apr 13, 2005 #5
    did u miss [tex](1-u)^{1/2}[/tex] -> [tex] \int cos t (2 sin^2 t + 1) dt[/tex]? Or did i miss something
     
  7. Apr 13, 2005 #6

    dextercioby

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    Yeah,u missed the second cosine.One from the sqrt & the other from the change of variable...

    Daniel.
     
  8. Apr 13, 2005 #7
    ah that clears things up. thanks.
     
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