Show that the arc length of the ellipse (x²)/2 + y² = 1

Show that the arc length of the ellipse $\dfrac{x^2}{2} + y^2 = 1$ equals the arc length of one wave of the sinusoid $y = \sin x$.

We compute each arc length as an integral and match them via a substitution.

Step 1: Parametrize the ellipse

With semi-axes $\sqrt2$ and $1$: $x=\sqrt2\cos t,\ y=\sin t,\ t\in[0,2\pi]$.

Step 2: Ellipse speed

\[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 = 2\sin^2t+\cos^2t = \sin^2t+(\sin^2t+\cos^2t) = 1+\sin^2t\]

\[L_{\text{ellipse}}=\int_0^{2\pi}\sqrt{1+\sin^2t}\,dt\]

Step 3: Sinusoid arc length

For $y=\sin x$, $y'=\cos x$, so one full wave ($x\in[0,2\pi]$) has

\[L_{\text{sine}}=\int_0^{2\pi}\sqrt{1+\cos^2x}\,dx\]

Step 4: Phase-shift the ellipse integral

Let $u=t-\frac\pi2$, so $\sin t=\cos u$; the limits become $u\in\left[-\frac\pi2,\frac{3\pi}2\right]$:

\[L_{\text{ellipse}}=\int_{-\pi/2}^{3\pi/2}\sqrt{1+\cos^2u}\,du\]

Step 5: Use periodicity

The integrand $\sqrt{1+\cos^2u}$ has period $\pi$ (since $\cos^2u$ does). Integrating it over any interval of length $2\pi$ — exactly two full periods — gives the same value regardless of where the interval starts, so

\[\int_{-\pi/2}^{3\pi/2}\sqrt{1+\cos^2u}\,du = \int_0^{2\pi}\sqrt{1+\cos^2u}\,du\]

Step 6: Conclude

The right-hand side is exactly $L_{\text{sine}}$ (the integration variable name does not matter), so

\[\boxed{L_{\text{ellipse}}=L_{\text{sine}}}\]

Numerically, both integrals evaluate to $L\approx7.640$ (confirmed by independent numerical integration of both forms), though the common value has no elementary closed form.