# Integration involving multiple constants…

So I've been tackling the rather nasty integral of...

$\int^R_s\frac{2r}{\sqrt{r^2-s^2}}.\frac{1}{2}(R-r)^2dr$

...where R and s are constants.

However, every method I try I seem to get stumped by something too awkward to integrate (for me at least).

My current method has been by integration by parts.

$u = \frac{1}{2}(R-r)^2, dv = \frac{2r}{\sqrt{r^2-s^2}}$

$du = -(R-r)dr, v = 2\sqrt{r^2-s^2}$

(I got v by using substition of $p = r^2-s^2, dp = 2r.dr$)

So integration by parts, $[uv]^R_s - \int^R_s v.du$

$uv = (R-r)^2\sqrt{r^2-s^2}$

$\int^R_sv.du = -2\int^R_s(R-r)\sqrt{r^2-s^2}dr = 2\int^R_sr\sqrt{r^2-s^2}dr-2R\int^R_s\sqrt{r^2-s^2}dr$

If we take the first integral...

$2\int^R_sr\sqrt{r^2-s^2}dr$, Let $p = r^2-s^2, dp = 2rdr$

=$\int^R_s\sqrt{p.}dp$

= $[\frac{2u^{\frac{3}{2}}}{3}]^R_s$

= $[\frac{2}{3}(r^2-s^2)^{\frac{3}{2}}]^R_s$

Now the second integral is the part I'm stuck on...

$2R\int^R_s\sqrt{r^2-s^2}dr$

I've tried various substitutions, messing around with sec and tan, and I just can't crack it.

Does anyone have any pointers? (Considering what I've done up to this point is correct, I just need this last integral before I can plob my results back into the integration by parts forumla and work out the definite integrals, etc.)

EDIT: I have tried substituting $r = sSecθ$, and came to the integral of $2Rs^2\int secθtan^2θdθ$

I'm having trouble integrating this too, as I keep ending back at the same integral.

• $r=s\sec\theta$ or $r=s\cosh t$ (and there are others). – André Nicolas Mar 10 '14 at 18:32
• @AndréNicolas I've tried sSecθ but I got lost with it. But I imagine it might lead to the right answer. As for hyperbolic functions, we haven't even touched on those before, so it's not something I can really work with at the moment. – Wolff Mar 10 '14 at 18:34
• OK, let's use $\sec$, either immediately or after your integrations (which I have not checked). When we substitute, we get, apart from constants, $\int \sec \theta \tan^2\theta\,d\theta$. Equivalently, we want to integrate $\sec\theta-\sec^3\theta$. This is unpleasant. There are "magic" ways, or else we can write $\sec\theta$ as $\frac{\cos\theta}{\cos^2\theta}=\frac{\cos\theta}{1-\sin^2\theta}$, then let $u=\sin\theta$ and use partial fractions. Easier is to look up or remember the integral of $\sec\theta$, or even of $\sqrt{w^2-1}$. – André Nicolas Mar 10 '14 at 18:49

By expanding the $(R-r)^2$ piece, you see that you basically are looking to evaluate

$$\int_s^R dr \frac{r^k}{\sqrt{r^2-s^2}}$$

for $k=1,2,3$. For $k=1$, the integral is simple:

$$\int_s^R dr \frac{r}{\sqrt{r^2-s^2}} = \frac12 \int_{s^2}^{R^2} \frac{du}{\sqrt{u-s^2}} = \sqrt{R^2-s^2}$$

For $k=2$, integration by parts is useful:

\begin{align}\int_s^R dr \frac{r^2}{\sqrt{r^2-s^2}} &= \left [r \sqrt{r^2-s^2} \right ]_s^R - \int_s^R dr \, \sqrt{r^2-s^2}\\ &= R\sqrt{R^2-s^2} - s^2 \int_0^{\cosh^{-1}{R/s}}dt \, \sinh^2{t}\\ &= R\sqrt{R^2-s^2} -\frac12 s^2\int_0^{\cosh^{-1}{R/s}}dt \,(\cosh{2 t}-1)\\ &= \frac12 R\sqrt{R^2-s^2} + \frac12 s^2\cosh^{-1}{\frac{R}{s}}\\ &= \frac12 R\sqrt{R^2-s^2} +\frac12 s^2\log{\left (\frac{R}{s} + \sqrt{\frac{R^2}{s^2}-1} \right )} \end{align}

For $k=3$, sub and integration by parts again:

\begin{align}\int_s^R dr \frac{r^3}{\sqrt{r^2-s^2}} &= \frac12 \int_{s^2}^{R^2} du \frac{u}{\sqrt{u-s^2}}\\ &= \frac12 \left [u \sqrt{u-s^2} \right ]_{s^2}^{R^2} - \frac12 \int_{s^2}^{R^2} du \, \sqrt{u-s^2}\\ &= \frac12 R^2\sqrt{R^2-s^2} - \frac13 \left (R^2-s^2 \right )^{3/2}\end{align}

Now put these pieces altogether from the original integral and, voila!