How to evaluate $I=\int_0^\pi\frac{\cos(\theta+\alpha)\sin(\theta)d\theta}{\sqrt{r^2+a^2-2ra\cos(\theta)}}$ Let's assume that $r>a>0$ and $0<\alpha<\pi$, to evaluate the following integral:
$$I=\int_0^\pi\frac{\cos(\theta+\alpha)\sin(\theta)d\theta}{\sqrt{r^2+a^2-2ra\cos(\theta)}}$$
I used a change of variable $u=\sqrt{r^2+a^2-2ra\cos(\theta)}$ to get:
$$I=\frac{1}{4r^2a^2}\int_{r-a}^{r+a}(\cos\alpha(r^2+a^2-u^2)-\sin\alpha \sqrt{(2ra)^2-(r^2+a^2-u^2)^2})du$$
Then I was stuck to integrate the complicated square root. Do you have an idea?
 A: For $\cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha$
$$
I = \cos\alpha \int_{0}^{\pi} \frac{\cos\theta\sin\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta - \sin\alpha \int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta
$$
The first integral is trivial, here I focus on the second
$$
\begin{aligned}
\int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta
&= \int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{(r+a)^2-4ra\cos^2\frac{\theta}{2}}} \,\mathrm{d}\theta \quad \text{let }k^2=\frac{4ra}{(r+a)^2} \text{ and } \frac{\theta}{2}\mapsto\theta \\
&= \frac2{r+a} \int_{0}^{\pi/2} \frac{\sin^22\theta}{\sqrt{1-k^2\cos^2\theta}} \,\mathrm{d}\theta \quad \theta\mapsto\frac{\pi}{2}-\theta \\
&= \frac2{r+a} \int_{0}^{\pi/2} \frac{\sin^22\theta}{\sqrt{1-k^2\sin^2\theta}} \,\mathrm{d}\theta
\end{aligned}
$$
Integrating by parts gives
$$
\int_{0}^{\pi/2}  \frac{\sin^22\theta}{\sqrt{1-k^2\sin^2\theta}} \,\mathrm{d}\theta = \frac4{k^2} \int_{0}^{\pi/2} \cos2\theta \sqrt{1-k^2\sin^2\theta} \,\mathrm{d}\theta
$$
where denote $I=\frac4{k^2}J$.
From another aspect, setting
$$
\sin^22\theta=A\cos2\theta(1-k^2\sin^2\theta) + B(1-k^2\sin^2\theta) + C
$$
or
$$
-4\sin^4\theta+4\sin^2\theta = 2k^2A\sin^4\theta - ((k^2+2)A+k^2B)\sin^2\theta + (A+B+C)
$$
comparing the coefficients, we solve
$$
A=-\frac2{k^2}, \quad B=-\frac{2(k^2-2)}{k^4}, \quad C=\frac{4(k^2-1)}{k^4}
$$
which suggests
$$
I=-\frac2{k^2}J + \frac{4(k^2-1)}{k^4}K(k^2) - \frac{2(k^2-2)}{k^4}E(k^2)
$$
Item of $J$ can be cancelled with $I=\frac4{k^2}J$, hence
$$
I = \int_{0}^{\pi/2}  \frac{\sin^22\theta}{\sqrt{1-k^2\sin^2\theta}} \,\mathrm{d}\theta = \frac{8(k^2-1)}{3k^4}K(k^2) - \frac{4(k^2-2)}{3k^4}E(k^2)
$$
or plug-in $k^2$
$$
\int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta = -\frac{r+a}{3r^2a^2}\left((r-a)^2K\left(\frac{4ra}{(r+a)^2}\right) - (r^2+a^2)E\left(\frac{4ra}{(r+a)^2}\right)\right)
$$
the rest of calculation is obvious.
A: First, simplify the expression writing
$${\sqrt{r^2+a^2-2ra\cos(\theta)}}=\sqrt{r^2+a^2} \sqrt{1-k \cos(\theta)}\quad \text{where} \quad k=\frac {2ra}{r^2+a^2}$$
There is even an antiderivative which contains a bunch of elliptic integrals of the first and second kind.
Using @Andrei suggestion, the numerator write
$$\sin (\alpha ) \cos ^2(\theta )+\cos (\alpha ) \sin (\theta ) \cos (\theta )-\sin
   (\alpha )$$ and the substitution $x=\cos(\theta)$ leads to "simple" integrals, the most difficult being the one with $\cos ^2(\theta )$ (but perfectly doable).
If I did not make any mistake, for the definite integral, we should arrive at
$$\color{blue}{\frac{2 a^3 \cos (\alpha )-(a+r) \sin (\alpha ) \left(\left(a^2+r^2\right)
   E\left(\frac{4 a r}{(a+r)^2}\right)-(a-r)^2 K\left(\frac{4 a
   r}{(a+r)^2}\right)\right)}{3 a^2 r^2}}$$
I just made a numerical check with $a=1$, $r=2$, $\alpha=\frac \pi 6$ which gives
$$\frac{1}{24} \left(3 K\left(\frac{8}{9}\right)+2 \sqrt{3}-15
   E\left(\frac{8}{9}\right)\right)=-0.235672$$ which is the value given by numerical integration.
