# Integral help

#### The_Duck

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.

#### paulfr

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

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 $u= R^2+ a^2- 2aR cos(\theta)$ so that $du= 2aR sin(\theta) d\theta$.

When $\theta= 0$, $u= R^2- 2aR+ a^2= (R- a)^2$.

When $\theta= \pi$, $u= R^2+ 2aR+ a^2= (R+ a)^2$.

$$\int_{(R-a)^2}^{(R+a)^2} u^{-3/2} du$$
$$= -\frac{2}{3}u^{-1/2}\right|_{(R- a)^2}^{(R+a)^2}$$
$$= \frac{2}{3}\left(\frac{1}{R-a}- \frac{1}{R+ a}\right)$$
$$= \frac{4}{3}\frac{a}{R^2- a^2}$$

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