Find $\int_a^b \sin |x| \, \mathrm{d}x $

Question

How to find the integral $$\int_a^b \sin |x| \, \mathrm{d}x \,?$$

I'm able to obtain definite integral of form $ \int_a^b \lvert\sin x \rvert \, \mathrm{d}x$ but not when the modulus operator is solely on $x$.

Answer 1

Hint: Look at the graph of $\sin x$ and $\sin |x|$ and see how the areas under each relate to each other.

Hint-er: Split the integral into two parts. $$\int_0^a \sin |x| \, \mathrm{d}x + \int_0^b \sin |x| \, \mathrm{d}x$$

What can you say about each of these integrals?

For $x>0$, the red graph overlays the blue one where red is $\sin |x|$ and blue is $\sin x$.

MORE HINTS: $$\int_0^b \sin |x| \, \mathrm{d}x = \int_0^b \sin x \, \mathrm{d}x$$

Answer 2

All antiderivatives of $\sin|x|$ have the form $$ c+\int_0^x\sin|t|\,dt $$ If $x\ge0$, we have $$ \int_0^x\sin|t|\,dt=\int_0^x\sin t\,dt=1-\cos x $$ whereas, if $x\le0$, $$ \int_0^x\sin|t|\,dt=\int_0^x\sin(-t)\,dt=-1+\cos(-x) $$ so the general antiderivative is $$ c+f(x) $$ where $$ f(x)=\begin{cases} 1-\cos x & \text{if $x\ge0$}\\ -1+\cos(-x) & \text{if $x<0$} \end{cases} $$

---

For a definite integral, there is no problem if $a<b\le0$ or $0\le a<b$. For the case $a\le 0\le b$, just split it as $$ \int_a^0\sin(-x)\,dx+\int_0^b\sin t\,dt $$ which is easier than using the antiderivative shown above.

Answer 3

Consider $$ f(x)=\left\{\begin{array}{cl} \dfrac{x}{|x|}(1-\cos(x))&\text{if }x\ne0\\ 0&\text{if }x=0\end{array}\right. $$ If $x\gt0$, then $f'(x)=\sin(x)$. If $x\lt0$, then $f'(x)=-\sin(x)$. The mean value theorem says that $f'(0)=0$. Therefore, $$ f'(x)=\sin(|x|) $$

Answer 4

$\because\int\sin|x|~dx$

$=\int\sin(x~\text{sgn}(x))~dx$

$=\text{sgn}(x)\int\sin x~dx$

$=\text{sgn}(x)\int_0^x\sin x~dx+C$

$=\text{sgn}(x)[-\cos x]_0^x+C$

$=\text{sgn}(x)(1-\cos x)+C$

$\therefore\int_a^b\sin|x|~dx$

$=[\text{sgn}(x)(1-\cos x)]_a^b$

$=\text{sgn}(b)(1-\cos b)-\text{sgn}(a)(1-\cos a)$