Given integral from 0 to (pi/2) sin^2nphi d
Given $\int_0^{\pi/2}\sin^{2n}\varphi\,d\varphi=\dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac\pi2$, show the arc length of the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}9=1$ equals $$ s = 10\pi\left\{1-\sum_{n=1}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\right]^2\frac{(4/5)^{2n}}{2n-1}\right\}. $$
Explanation of Method
Parametrizing the ellipse and simplifying the arc-length element leads to $a\sqrt{1-e^2\sin^2\varphi}$, whose binomial expansion, integrated term by term using the given Wallis-type formula, reproduces the classical series for the elliptic integral of the second kind.
- Simplify the arc-length element into the eccentricity form $a\sqrt{1-e^2\sin^2\varphi}$.
- Expand the square root via the binomial series and derive its coefficients explicitly.
- Integrate term by term using the given Wallis formula and simplify to the target form.
Step 1 · Arc-length element in eccentricity form
With $a=5$, $b=3$, and $x=a\cos\theta$, $y=b\sin\theta$: $$ ds = \sqrt{a^2\sin^2\theta+b^2\cos^2\theta}\,d\theta = a\sqrt{1-e^2\cos^2\theta}\,d\theta, \qquad e^2=\frac{a^2-b^2}{a^2}=\frac{16}{25}. $$ Substituting $\varphi=\tfrac\pi2-\theta$ turns $\cos\theta$ into $\sin\varphi$, so by symmetry the total arc length is $$ s = 4a\int_0^{\pi/2}\sqrt{1-e^2\sin^2\varphi}\,d\varphi = 20\int_0^{\pi/2}\sqrt{1-e^2\sin^2\varphi}\,d\varphi. $$
Step 2 · Binomial expansion of the square root
Using $\binom{1/2}n=\dfrac{(-1)^{n-1}(2n-3)!!}{2^nn!}$ for $n\ge1$, and substituting $t=-e^2\sin^2\varphi$: $$ \sqrt{1-e^2\sin^2\varphi} = 1 - \sum_{n=1}^{\infty}\frac{(2n-3)!!}{2^nn!}e^{2n}\sin^{2n}\varphi. $$
Step 3 · Integrate and simplify
Integrating term by term with the given Wallis formula: $$ \int_0^{\pi/2}\!\!\sqrt{1-e^2\sin^2\varphi}\,d\varphi = \frac\pi2\left[1-\sum_{n=1}^{\infty}\frac{(2n-3)!!}{2^nn!}\cdot\frac{(2n-1)!!}{(2n)!!}\,e^{2n}\right]. $$ Since $(2n)!!=2^nn!$, the coefficient simplifies: $$ \frac{(2n-3)!!(2n-1)!!}{[(2n)!!]^2} = \frac{(2n-1)[(2n-3)!!]^2}{[(2n)!!]^2} = \frac1{2n-1}\left[\frac{(2n-1)!!}{(2n)!!}\right]^2, $$ using $(2n-1)!!=(2n-1)(2n-3)!!$. Therefore $$ s = 20\cdot\frac\pi2\left\{1-\sum_{n=1}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\right]^2\frac{e^{2n}}{2n-1}\right\} = 10\pi\left\{1-\sum_{n=1}^{\infty}\left[\frac{(2n-1)!!}{(2n)!!}\right]^2\frac{(4/5)^{2n}}{2n-1}\right\}. $$
Summary Table
| Property | Value |
|---|---|
| Eccentricity | $e=4/5$ |
| Series | $10\pi\{1-\sum[(2n-1)!!/(2n)!!]^2(4/5)^{2n}/(2n-1)\}$ |
| Numerical value | $s\approx25.53$ |
Additional Notes
- This series is exactly $4a\,E(e)$, where $E(e)=\int_0^{\pi/2}\sqrt{1-e^2\sin^2\varphi}\,d\varphi$ is the classical complete elliptic integral of the second kind — the arc length of any ellipse reduces to this same series, only $a$ and $e$ change.
- Numerically, $s\approx25.53$, matching Ramanujan's well-known approximation $\pi[3(a+b)-\sqrt{(3a+b)(a+3b)}]\approx25.53$ closely — a useful independent check since elliptic integrals have no elementary closed form.
- Since $e=4/5<1$ strictly, $(4/5)^{2n}\to0$ geometrically, guaranteeing convergence; the closer the ellipse is to a circle ($e\to0$), the faster this series converges, while a highly eccentric ellipse ($e\to1$) converges only very slowly.