$\int_{-\pi / 2}^{\pi / 2} \sin |x| d x$ is equal to
$\int_{-\pi / 2}^{\pi / 2} \sin |x| d x$ is equal to
$0$
$1$
$2$
$\pi$
Solution
Let
$
\begin{aligned}
I & =\int_{-\pi / 2}^{\pi / 2} \sin |x| d x \\
& =2 \int_0^{\pi / 2} \sin x d x \\
& =2[-\cos x]_0^{\pi / 2} \\
& =2
\end{aligned}
$