# How find the integral $I=\int_{-R}^{R}\frac{\sqrt{R^2-x^2}}{(a-x)\sqrt{R^2+a^2-2ax}}dx$

Find the integral:

$$I=\int_{-R}^{R}\dfrac{\sqrt{R^2-x^2}}{(a-x)\sqrt{R^2+a^2-2ax}}\;\mathrm dx$$

My try:

Let $x=R\sin{t},\;t\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ then, $$I=\int_{-\pi/2}^{\pi/2}\dfrac{R\cos{t}}{(a-R\sin{t})\sqrt{R^2+a^2-2aR\sin{t}}}\cdot R\cos{t}\;\mathrm dt$$ so, $$I=R^2\int_{-\pi/2}^{\pi/2}\dfrac{\cos^2{t}}{(a-R\sin{t})\sqrt{R^2+a^2-2aR\sin{t}}}\;\mathrm dt$$

Maybe following can use Gamma function? But I can't find it. Thank you someone can help me.

• Wouldn't it help to first rescale the integral down to one parameter? Substituting $x=Ru$, I reduced the integral to $I=\int_{-1}^{1}\dfrac{\sqrt{1-u^2}}{(\alpha-u)\sqrt{1+\alpha^2-2\alpha u}}du$. Just seems like a cleaner place to start to me. – David H Oct 22 '13 at 15:35

## 1 Answer

The second line of the OP is $$I=\int_{-\pi/2}^{\pi/2} \frac{\cos^2 t} {(\alpha -\sin t)\sqrt{1+\alpha^2-2\alpha \sin t}}dt$$ $$=\int_{-\pi/2}^{\pi/2} \frac{1-\sin^2 t} {(\alpha -\sin t)\sqrt{1+\alpha^2-2\alpha \sin t}}dt$$ where $\alpha\equiv a/R$. Partial fraction decomposition yields $$=\int_{-\pi/2}^{\pi/2} [\sin t +\alpha+\frac{1-\alpha^2}{\alpha -\sin t}]\frac{1} {\sqrt{1+\alpha^2-2\alpha \sin t}}dt.$$ Then the three integrals $$\int \frac{1}{\sqrt{a+b\sin x}}dx,$$ $$\int \frac{\sin x}{\sqrt{a+b\sin x}}dx,$$ and $$\int \frac{1}{(2-p^2+p^2\sin x)\sqrt{a+b\sin t}}dt$$ are tabulated in terms of Elliptic Integrals as 2.571.1, 2.571.2 and 2.574.1 in the Gradsteyn-Rzyshik tables of integrals.