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Integration using trig subs having trouble switching back to the original variable

  1. Oct 4, 2004 #1
    Hi, I carried out the integration until the very end...I don't know how to convert the variable back to the original one. :confused:

    [tex]\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^2-u^2}} [/tex]

    [tex] Let u = a\sin{\Theta}[/tex]
    [tex] du = a\cos{\Theta}d\Theta[/tex]

    The integral becomes...

    [tex]\int_{R_{0}}^{R(\Theta)}\frac{a\cos{\Theta}d\Theta}{a\cos{\Theta}a\sin{\Theta}}[/tex]
    [tex]\frac{1}{a}\int_{R_{0}}^{R(\Theta)}\csc{\Theta}d{\Theta}[/tex]

    [tex]\csc{\Theta}d{\Theta} = \ln {|\csc{\Theta}-\cot{\Theta}|}+C[/tex]

    This is where I'm stuck. I don't know how to convert the thetas back into the "u"s. I haven't multiplied the answer by 1/a yet. I know that [tex]\Theta=\sin^{-1}{u/a}[/tex], but if I plug the [tex]\sin^{-1}{u/a}[/tex] into Theta, the expression becomes super messy and I really don't know what to do with it.

    Please help, thanks in advance! :smile:
     
    Last edited: Oct 4, 2004
  2. jcsd
  3. Oct 4, 2004 #2
    make a triangle.. it's the only way. For example, if theta = arcsin(u/a) then sin(arcsin(u/a)) = u/a, cos(arcsin(u/a)) = sqrt(a^2-u^2)/a. Make a triangle with sides u, sqrt(a^2-u^2) and a. You can find all the trig functions from it (be sure to label theta)
     
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