Janet Forbes

Answered

2022-07-02

Antiderivative of $|\mathrm{sin}(x)|$

Because ${\int}_{0}^{\pi}\mathrm{sin}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=2,$, then ${\int}_{0}^{16\pi}|\mathrm{sin}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=32.$.

And Wolfram Alpha agrees to this, but when I ask for the indefinite integral $\int |\mathrm{sin}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x,$,

Wolfram gives me $-\mathrm{cos}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{g}\mathrm{n}(\mathrm{sin}(x))+c.$.

However, $[-\mathrm{cos}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{g}\mathrm{n}(\mathrm{sin}(x)){]}_{0}^{16\pi}=0,$,

So what's going on here? What is the antiderivative of $|\mathrm{sin}(x)|$?

Because ${\int}_{0}^{\pi}\mathrm{sin}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=2,$, then ${\int}_{0}^{16\pi}|\mathrm{sin}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=32.$.

And Wolfram Alpha agrees to this, but when I ask for the indefinite integral $\int |\mathrm{sin}(x)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x,$,

Wolfram gives me $-\mathrm{cos}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{g}\mathrm{n}(\mathrm{sin}(x))+c.$.

However, $[-\mathrm{cos}(x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{s}\mathrm{g}\mathrm{n}(\mathrm{sin}(x)){]}_{0}^{16\pi}=0,$,

So what's going on here? What is the antiderivative of $|\mathrm{sin}(x)|$?

Answer & Explanation

Mateo Carson

Expert

2022-07-03Added 15 answers

Step 1

Let $f(x)={\int}_{0}^{x}|\mathrm{sin}x|\phantom{\rule{thinmathspace}{0ex}}dx=(1-\mathrm{cos}x),0\le x\le \pi .$.

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 $

Step 2

And so on. you can verify that $f(n\pi )=2n\text{for all integer}n.$.

in particular $f(16\pi )=32.$.

Let $f(x)={\int}_{0}^{x}|\mathrm{sin}x|\phantom{\rule{thinmathspace}{0ex}}dx=(1-\mathrm{cos}x),0\le x\le \pi .$.

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 $

Step 2

And so on. you can verify that $f(n\pi )=2n\text{for all integer}n.$.

in particular $f(16\pi )=32.$.

Savanah Boone

Expert

2022-07-04Added 5 answers

Step 1

For $x\in [n\pi ,(n+1)\pi ]$, we get

$\begin{array}{}\text{(1)}& {\int}_{0}^{n\pi}|\mathrm{sin}(t)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t=2n\end{array}$

and $\begin{array}{rl}{\int}_{n\pi}^{x}|\mathrm{sin}(t)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t& ={\int}_{0}^{x-n\pi}\mathrm{sin}(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t\\ \text{(2)}& & =1-\mathrm{cos}(x-n\pi )\end{array}$

Step 2

Piecing (1) and (2) together yields $\begin{array}{}\text{(3)}& {\int}_{0}^{x}|\mathrm{sin}(t)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t=1-\mathrm{cos}(x-\pi \stackrel{n}{\stackrel{\u23de}{\lfloor x/\pi \rfloor}})+2\stackrel{n}{\stackrel{\u23de}{\lfloor x/\pi \rfloor}}\end{array}$

For $x\in [n\pi ,(n+1)\pi ]$, we get

$\begin{array}{}\text{(1)}& {\int}_{0}^{n\pi}|\mathrm{sin}(t)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t=2n\end{array}$

and $\begin{array}{rl}{\int}_{n\pi}^{x}|\mathrm{sin}(t)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t& ={\int}_{0}^{x-n\pi}\mathrm{sin}(t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t\\ \text{(2)}& & =1-\mathrm{cos}(x-n\pi )\end{array}$

Step 2

Piecing (1) and (2) together yields $\begin{array}{}\text{(3)}& {\int}_{0}^{x}|\mathrm{sin}(t)|\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t=1-\mathrm{cos}(x-\pi \stackrel{n}{\stackrel{\u23de}{\lfloor x/\pi \rfloor}})+2\stackrel{n}{\stackrel{\u23de}{\lfloor x/\pi \rfloor}}\end{array}$

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