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Homework Help: Math Physics Tricky Integrals go Over my Head

  1. Feb 20, 2008 #1
    so i have this question about a ladder and the area under it and when it all comes down to it I get the integral, ∫ u2 ( 1 – u2 ) 3/2 du, from u = 0 to u = 1.


    I am not sure whether to use trig substitution. I keep ending up in the same boat when I do. . . And the same thing when I do integration by parts. I'm not too quick with integration obviously. any help?
     
  2. jcsd
  3. Feb 20, 2008 #2
    is u2 u*2?

    I'm almost gonna have to guess it's u*(1-u)^(-3/2) that's giving you trouble(or 2*u or whatever)

    in which case I think integration by parts will do it

    What's the actual problem though?
     
  4. Feb 20, 2008 #3

    clem

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    I assume you mean [tex]\int_0^1 u^2(1-u^2)^{3/2} du[/tex].
    Let [tex]u=\sin\theta[/tex] and brush up on your trig.
    Use the formulae for sin and cos of theta/2.
     
  5. Feb 20, 2008 #4
    oh ok, that makes more sense with what he typed >_>

    You'll need those trig identities, but you can find huge tables of such identities in moments courtesy of the internet
     
  6. Feb 20, 2008 #5
    I tried substituting u = sin(theta) and all I got was sin^2(theta) cos^4(theta) dtheta... and that just doesn't mean anything to me...
     
  7. Feb 20, 2008 #6

    dextercioby

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    [tex] \int \sin^2 \theta {} \cos^4 \theta {} {} d\theta = \frac{1}{4}\int \sin^2 2\theta \ \frac{1+\cos 2\theta}{2} {} d\theta [/tex]

    Can you carry on from here ?
     
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