Show that length of sine is equal to length of cosine on the same interval. Let $$f(x)=\sin(x)\\\ g(x)=\cos(x)$$
Let $L_1$ be $$\int_0^{2\pi}\sqrt{1+\cos^2(x)}\space dx$$
And $L_2$
$$\int_0^{2\pi}\sqrt{1+\sin^2(x)}\space dx$$
I.e. L is a length of sine/cosine during it's period interval.
Numerical approach shows that these two integrals are equal. It seems reasonable, as both have the same wavelength and frequency.
How can one show above relationship holds? ($L_1=L_2$)
 A: Basically, we want to exploit the fact that $\sin$ and $\cos$ "look the same". The theorem that lets us do this is integration by substitution. Let's define, for convenience, the functions $\ell_1(x) = \sqrt{1 + \cos^2 x}$, and $\ell_2(x) = \sqrt{1 + \sin^2 x}$, so
\begin{align*}
 L_1 &= \int_0^{2\pi} \ell_1(x) \,\mathrm dx \\
 L_2 &= \int_0^{2\pi} \ell_2(x) \,\mathrm dx
\end{align*}
Now note that
\begin{align*}
 L_1 = \int_0^{\pi/2} \ell_1(x) \,\mathrm dx
  + \int_{\pi/2}^{2\pi} \ell_1 (x) \,\mathrm dx \\
 L_2 = \int_0^{\pi/2} \ell_2(x) \,\mathrm dx
  + \int_{\pi/2}^{2\pi} \ell_2 (x) \,\mathrm dx
\end{align*}
However, using the substitution $u = \pi/2 - x$, as given in the comments,
\begin{align*}
 \int_0^{\pi/2} \ell_1(x) \,\mathrm dx
  &= \int_{\pi/2}^0 (-\ell_2(u)) \,\mathrm du \\
  &= \int_0^{\pi/2} \ell_2(u) \,\mathrm du
\end{align*}
And from the substitution $u = \tfrac 52 \pi - x$,
\begin{align*}
 \int_{\pi/2}^{2\pi} \ell_1 (x) \,\mathrm dx
  &= \int_{2\pi}^{\pi/2} (-\ell_2 (u)) \,\mathrm du \\
  &= \int_{\pi/2}^{2\pi} \ell_2 (u) \,\mathrm du
\end{align*}
It is an exercise in trigonometric identities to establish that
$\ell_1(\pi/2 - u) = \ell_2(u)$, and $\ell_2(\tfrac 52 \pi - u) = \ell_2(u)$.
From these equalities, it follows that $L_1 = L_2$.
In fact, you can use this approach to show that for any function $f$,
\begin{equation*}
 \int_0^{2\pi} f(\sin x) \,\mathrm dx =
 \int_0^{2\pi} f(\cos x) \,\mathrm dx
\end{equation*}
This also holds on smaller intervals like $[0, \tfrac 12 \pi]$ or $[0, \pi]$.
A: Use property of definite integration: $\int_0^{2a}f(x)dx=2\int_0^a f(x)dx\ \  \forall\ \ f(2a-x)=f(x)$
$$L_1=\int_0^{2\pi}\sqrt{1+\cos^2(x)}\ dx=2\int_0^{\pi}\sqrt{1+\cos^2(x)}\ dx=4\int_0^{\pi/2}\sqrt{1+\cos^2(x)}\ dx$$
$$L_2=\int_0^{2\pi}\sqrt{1+\sin^2(x)}\ dx=2\int_0^{\pi}\sqrt{1+\sin^2(x)}\ dx=4\int_0^{\pi/2}\sqrt{1+\sin^2(x)}\ dx$$
Use property of definite integration: $\int_a^{b}f(x)dx=\int_a^b f(a+b-x)dx$
$$=4\int_0^{\pi/2}\sqrt{1+\sin^2\left(\frac{\pi}{2}-x\right)}\ dx=4\int_0^{\pi/2}\sqrt{1+\cos^2(x)}\ dx$$
$$\therefore L_1=L_2$$
