Antiderivative of $|\sin(x)|$

Question

Because $$\int_0^{\pi}\sin(x)\,\mathrm{d}x=2,$$ then $$\int_0^{16\pi}|\sin(x)|\,\mathrm{d}x=32.$$ And Wolfram Alpha agrees to this, but when I ask for the indefinite integral $$\int|\sin(x)|\,\mathrm{d}x,$$ Wolfram gives me $$-\cos(x)\,\mathrm{sgn}(\sin(x))+c.$$ However, $$[-\cos(x)\,\mathrm{sgn}(\sin(x))]_0^{16\pi}=0,$$ So what's going on here? What is the antiderivative of $|\sin(x)|$?

Answer 1

For $x\in[n\pi,(n+1)\pi]$, we get $$ \int_0^{n\pi}|\sin(t)|\,\mathrm{d}t=2n\tag{1} $$ and $$ \begin{align} \int_{n\pi}^x|\sin(t)|\,\mathrm{d}t &=\int_0^{x-n\pi}\sin(t)\,\mathrm{d}t\\ &=1-\cos(x-n\pi)\tag{2} \end{align} $$ Piecing $(1)$ and $(2)$ together yields $$ \int_0^x|\sin(t)|\,\mathrm{d}t=1-\cos(x-\pi\overbrace{\lfloor x/\pi\rfloor}^n)+2\overbrace{\lfloor x/\pi\rfloor}^n\tag{3} $$

Answer 2

let $$f(x) = \int_0^x |\sin x| \, dx = (1-\cos x), 0\le x \le \pi.$$ since the integrand is $\pi$-periodic, we can extend the formula for $$f(x) = f(\pi) + f(x-\pi), \pi \le x \le 2\pi$$ and so on. you can verify that $$f(n\pi) = 2n \text{ for all integer } n.$$ in particular $f(16\pi) = 32.$

Answer 3

Using symmetry : $\int_0^\pi|\sin x|dx=2$ then $\int_0^{16\pi}|\sin x|dx=16.2=32$

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \begin{align}&\color{#66f}{\large\int\verts{\sin\pars{x}}\,\dd x} =x\verts{\sin\pars{x}}-\int x\,{\rm sgn}\pars{\sin\pars{x}}\cos\pars{x}\,\dd x \\[5mm]&=x\verts{\sin\pars{x}} -\int\,{\rm sgn}\pars{\sin\pars{x}}\,\dd\bracks{\cos\pars{x} + x\sin\pars{x}} \\[1cm]&=x\verts{\sin\pars{x}} -\bracks{\cos\pars{x} + x\sin\pars{x}}\,{\rm sgn}\pars{\sin\pars{x}} \\[5mm]&+\int\bracks{\cos\pars{x} + x\sin\pars{x}}2\delta\pars{\sin\pars{x}} \cos\pars{x}\,\dd x \\[1cm]&=-\cos\pars{x}\,{\rm sgn}\pars{\sin\pars{x}} +2\int\delta\pars{\sin\pars{x}}\,\dd x \\[5mm]&=\color{#66f}{\large -\cos\pars{x}\,{\rm sgn}\pars{\sin\pars{x}} +2\sum_{n=-\infty}^{\infty}\ \int\delta\pars{x - n\pi}\,\dd x} \end{align}

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Then, $$ \int_{0}^{16\pi}\verts{\sin\pars{x}}\,\dd x =\overbrace{\left.\vphantom{\LARGE A}-\cos\pars{x}\,{\rm sgn}\pars{\sin\pars{x}} \right\vert_{\ 0}^{\ 16\pi}}^{\ds{=\ \dsc{0}}}\ +\ 2\ \overbrace{\sum_{n=-\infty}^{\infty}\ \int_{0^{-}}^{\pars{16\pi}^{\, +}}\delta\pars{x - n\pi}\,\dd x}^{\ds{=\ \dsc{16}}} \ =\ 32 $$

Answer 5

The $c$ is important here! This is a subtle issue that comes up with formulas for antiderivatives: at any point where the antiderivative "without the $c$" is discontinuous, the value of $c$ can change.

Suppose $I$ is an interval between two consecutive roots of $\sin(x)$.

• If $\sin(x)$ is positive on $I$, then $|\sin(x)| = \sin(x)$ and its antiderivative is $-\cos(x) + c$.

• If $\sin(x)$ is negative on $I$, then $|\sin(x)| = -\sin(x)$ and its antiderivative is $\cos(x) + c$.

So one antiderivative of $\sin(x)$ really is $$-\cos(x)\operatorname{sgn}(\sin(x)) + c,$$ at least on every such interval $I$. And of course the only points left out are the roots of $\sin(x)$, which form a discrete set. Here is a graph of that function, with $c = 0$, from Wolfram Alpha. As you can see. it has a jump discontinuity at each root of $\sin(x)$.

The reason that the integrals in the original post don't work out is that if we want an antiderivative that is defined on a region that is more than the interval between two roots, the $c$ must change at every root of $\sin(x)$ to give a continuous antiderivative. This is why the naive integral calculation done in the post is flawed - because $c$ is only constant on each interval $I$.

If you look at a graph of $-\cos(x)\operatorname{sgn}(\sin(x))$ above, you will see that it has a jump discontinuity of $2$ at each root (because $\int_0^\pi \sin(x) = 2$), and that it does "flatten out" at each root. So an antiderivative of $|\sin(x)|$ defined on all of $\mathbb{R}$ is $$-\cos(x)\operatorname{sgn}(\sin(x)) + j(x) + c,$$ where $j(x)$ is a particular step function that increases by $2$ at each root of $\sin(x)$. But, in a table of integrals,the $j(x)$ may seem to be "hidden" inside the $c$.

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We can look at another example. which is a little easier because it does not have any periodic nature. Consider $\int |e^x - 1|\,dx$. It is easy to work out an antiderivative $f(x)$ in $\mathbb{R} \setminus \{0\}$: $$ f(x) = \begin{cases} e^x - x + c & ; x > 0, \\ -e^x + x + c & ; x < 0. \end{cases} $$ We may be tempted to write this as $f(x) = (e^x - x)\operatorname{sgn}(e^x - 1) + c$, and that is correct on $\mathbb{R} \setminus \{0\}$, although the '$c$' can be different on each side of $0$.

Now, let's look at the graph of $f(x)$ (with $c = 0$ everywhere) from Wolfram Alpha. There is a jump discontinuity at $x = 0$.

You can see (and verify algebraically) that $$\lim_{x \to 0^-} f'(x) = 0 = \lim_{x \to 0^+} f'(x) = (e^x - 1)\big |_{x =0}.$$ So we can make $f$ continuous and differentiable on $\mathbb{R}$ by choosing $c$ appropriately on each side of $0$ to eliminate the jump discontinuity. The resulting function will be an antiderivative of $|e^x - 1|$ that is correct on all of $\mathbb{R}$.

Answer 6

According to http://en.wikipedia.org/wiki/Lists_of_integrals#Absolute-value_functions,

$\int|\sin x|~dx=2\left\lfloor\dfrac{x}{\pi}\right\rfloor-\cos{\left(x-\pi\left\lfloor\dfrac{x}{\pi}\right\rfloor\right)}+C$