6
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – David H Oct 22 '13 at 15:35
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.