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$\int_{-\pi / 2}^{\pi / 2} \sin |x| d x$ is equal to
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2098 Upvotes
Verified Answer
The correct answer is:
$2$
Let
$$
\begin{aligned}
I & =\int_{-\pi / 2}^{\pi / 2} \sin |x| d x \\
& =2 \int_0^{\pi / 2} \sin x d x \\
& =2[-\cos x]_0^{\pi / 2} \\
& =2
\end{aligned}
$$
$$
\begin{aligned}
I & =\int_{-\pi / 2}^{\pi / 2} \sin |x| d x \\
& =2 \int_0^{\pi / 2} \sin x d x \\
& =2[-\cos x]_0^{\pi / 2} \\
& =2
\end{aligned}
$$
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