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

Question

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?

Answer 1

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.

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

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

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

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

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