Calculating the integral $\int_{0}^{\pi /6}\sqrt{1-\left(\frac{R_s\sin \theta }{C_L}\right)^2} d\theta$

I want to integrate $I=\int\limits_{0}^{\pi /6}{\sqrt{1-{{\left( \frac{{{R}_{s}}\sin \theta }{{{C}_{L}}} \right)}^{2}}}d \theta}$. I get incomplete elliptic integral $E(z\mid m)$ in the calculation by mathematica. I need some simple calulation for including the function in further calulations along with other functions. Any way to proceed directly, without the help of Elliptic Integral?

• Since you can't simplify, why not keep it as it is in calculations?vand evaluate in the end using elliptic integrals. Alternatively, is there any problem in evaluating to greater significant digit than needed so that later calculations do not create loss of significant digits? – user45099 Apr 5 '13 at 11:49
• Sorry I had some typo-error in equation, correted it now. – Ashish Apr 5 '13 at 11:54
• @user57 thought to do so, but was just like carry forwarding my problem to the next stage. I am doing this calculation to get of overlap between two sections and I have no idea how to interpret $E\left(\frac{\pi }{3}|\frac{R^2}{C^2}\right)$ even if it is in final result to get the exact area. – Ashish Apr 5 '13 at 12:04
• "Any way to proceed directly, without the help of Elliptic Integral?" - no, but if the goal is numerical evaluation, there are efficient methods for numerically evaluating the elliptic integral of the second kind, based on the arithmetic-geometric mean. – J. M. is a poor mathematician Apr 5 '13 at 12:33
• First: a terminological note: elliptic integrals are not elliptic functions. They are inverses of elliptic functions (or at least the integral of the first kind is). Second: yes, the incomplete elliptic integral can also be computed via the AGM, but it takes more work precisely due to the incompleteness. But before all that: what are the parameter ranges for $R_S$ and $C_L$? – J. M. is a poor mathematician Apr 5 '13 at 12:51

For the binomial series of $\sqrt{1-x}$ , $\sqrt{1-x}=\sum\limits_{n=0}^\infty\dfrac{(2n)!x^n}{4^n(n!)^2(1-2n)}$

$\therefore\int_0^{\frac{\pi}{6}}\sqrt{1-\left(\dfrac{R_s\sin\theta}{C_L}\right)^2}~d\theta=\int_0^{\frac{\pi}{6}}\sum\limits_{n=0}^\infty\dfrac{(2n)!R_s^{2n}\sin^{2n}\theta}{4^n(n!)^2(1-2n)C_L^{2n}}d\theta=\int_0^{\frac{\pi}{6}}\left(1+\sum\limits_{n=1}^\infty\dfrac{(2n)!R_s^{2n}\sin^{2n}\theta}{4^n(n!)^2(1-2n)C_L^{2n}}\right)d\theta$

Now for $\int\sin^{2n}\theta~d\theta$ , where $n$ is any natural number,

$\int\sin^{2n}\theta~d\theta=\dfrac{(2n)!\theta}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}\theta\cos\theta}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts.

$\therefore\int_0^{\frac{\pi}{6}}\left(1+\sum\limits_{n=1}^\infty\dfrac{(2n)!R_s^{2n}\sin^{2n}\theta}{4^n(n!)^2(1-2n)C_L^{2n}}\right)d\theta$

$=\left[\theta+\sum\limits_{n=1}^\infty\dfrac{((2n)!)^2R_s^{2n}\theta}{4^{2n}(n!)^4(1-2n)C_L^{2n}}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2((k-1)!)^2R_s^{2n}\sin^{2k-1}\theta\cos\theta}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)C_L^{2n}}\right]_0^{\frac{\pi}{6}}$

$=\left[\sum\limits_{n=0}^\infty\dfrac{((2n)!)^2R_s^{2n}\theta}{4^{2n}(n!)^4(1-2n)C_L^{2n}}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2((k-1)!)^2R_s^{2n}\sin^{2k-1}\theta\cos\theta}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)C_L^{2n}}\right]_0^{\frac{\pi}{6}}$

$=\sum\limits_{n=0}^\infty\dfrac{((2n)!)^2R_s^{2n}\pi}{4^{2n}6(n!)^4(1-2n)C_L^{2n}}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2((k-1)!)^2R_s^{2n}\sin^{2k-1}\dfrac{\pi}{6}\cos\dfrac{\pi}{6}}{4^{2n-k+1}(n!)^4(2k-1)!(1-2n)C_L^{2n}}$

$=\sum\limits_{n=0}^\infty\dfrac{((2n)!)^2R_s^{2n}\pi}{2^{2n+1}3(n!)^4(1-2n)C_L^{2n}}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((2n)!)^2((k-1)!)^2R_s^{2n}\sqrt3}{4^{2n+1}(n!)^4(2k-1)!(1-2n)C_L^{2n}}$