$f(t) = \int_0^t |sin s| \ ds$, then $f(10\pi) = 5f(2\pi)$ I need to prove that if $f(t) = \int_0^t |sin s| \ ds$, then $f(10\pi) = 5f(2\pi)$, using the change of variable theorem. That's what I did:
$$f(10\pi) = \int_0^{10\pi} |\sin s| \ ds $$
take $s = 5x$, then $s^{-1}(0) = 0$ and $s^{-1}(10\pi) = 2\pi$, $ds = 5dx$, therefore
$$f(10\pi) = \int_0^{10\pi} |\sin s| \ ds = 5\int_0^{2\pi} |\sin 5x|dx$$
but I did not end with $5f(2\pi)$. What am I doing wrong?
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\verts{\sin\pars{s}}}$ is a periodic function of period $\ds{\pi}$:
\begin{align}
\color{#f00}{\int_{0}^{10\pi}\verts{\sin\pars{s}}\,\dd s} & =
10\int_{0}^{\pi}\verts{\sin\pars{s}}\,\dd s =
10\int_{0}^{\pi}\sin\pars{s}\,\dd s = \color{#f00}{20}
\end{align}
A: HINT:
Show that
$$\int_{2k\pi}^{2k\pi+2\pi}|\sin(s)|\,ds=\int_0^{2\pi}|\sin(s)|\,ds$$
for all integer $k$ by making a change of variable.
