How do I evaluate integrals that involve the signum ($\text{sgn}$) function?

Question

For example, I want to evaluate $$ \displaystyle \int_{0}^{2\pi} \left| \sin x \right| \text{ d}x $$ and I already know that: $$ \displaystyle \begin{aligned} \int \left| \sin x \right| \text{ d}x & = \int \sin x \text{ sgn}\left( \sin x \right) \text{ d}x \\ & = -\cos x \text{ sgn}\left( \sin x \right) + \mathcal{C} \end{aligned}$$ How would I evaluate the definite integral involving the signum function?

Answer 1

Hint: Instead of trying to use the sign function, break up the integral into intervals where $\sin x \geq 0$ and $\sin x < 0$. On such intervals, $|\sin x|$ is particularly simple and easier to integrate.

Answer 2

HINT : $$ \int_{0}^{\large 2\pi} \left| \sin x \right|\ dx=\int_{0}^{\large\pi}\sin x\ dx-\int_{\large\pi}^{\large2\pi}\sin x\ dx. $$

Answer 3

Let $x = u + \pi \to dx = du$, and $sinx = -sinu$. Thus:

$\displaystyle \int_{0}^{2\pi} |sinx|dx = \displaystyle \int_{-\pi}^{\pi} |sinu|du$.

Observe that $f(u) = |sinu|$ is an even function, so:

$\displaystyle \int_{-\pi}^{\pi} |sinu|du = 2\displaystyle \int_{0}^\pi |sinu|du = 2\displaystyle \int_{0}^\pi sinudu = (-2cosu)|_{u=0}^\pi = 4$.