Integral help

  • Thread starter Dawei
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Suppose you make the substitution u = cos(theta). I think that ought to simplify it a bit more.

Edit: I had sin(theta) earlier but I think you want cos.
 
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Yes u = cos theta and du = - sin(theta) d(theta)
yields ....... minus Integral (p + qu)^(-3/2) du
where ....... p = R^2 + a^2, q = -2Ra
which becomes
- 2 [ (p+qu)^(-1/2) ] + C
backsubstituting
- 2 [ (p+qcos(theta))^(-1/2) ] + C

I think I may be off by a constant factor here.
The (p+qu)^(-3/2) term needs to be modified to
k (p/q +u)^(-3/2) before integrating.
 
Last edited:

HallsofIvy

Science Advisor
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Just because you don't get a response immediately doesn't mean "no one can do it". In fact, that's a fairly elementary integral.

Let [itex]u= R^2+ a^2- 2aR cos(\theta)[/itex] so that [itex]du= 2aR sin(\theta) d\theta[/itex].

When [itex]\theta= 0[/itex], [itex]u= R^2- 2aR+ a^2= (R- a)^2[/itex].

When [itex]\theta= \pi[/itex], [itex]u= R^2+ 2aR+ a^2= (R+ a)^2[/itex].

Your integral becomes
[tex]\int_{(R-a)^2}^{(R+a)^2} u^{-3/2} du[/tex]
[tex]= -\frac{2}{3}u^{-1/2}\right|_{(R- a)^2}^{(R+a)^2}[/tex]
[tex]= \frac{2}{3}\left(\frac{1}{R-a}- \frac{1}{R+ a}\right)[/tex]
[tex]= \frac{4}{3}\frac{a}{R^2- a^2}[/tex]
 

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