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$\int_{-\pi / 2}^{\pi / 2} \sin x d x$
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Verified Answer
The correct answer is:
0
$$
\begin{aligned}
&\int_{-\pi / 2}^{\pi / 2} \sin x d x=\int_{-\pi / 2}^{0} \sin x d x+\int_{0}^{\pi / 2} \sin x d x \\
&=-[\cos x]_{-\pi / 2}^{0}-[\cos x]_{0}^{\pi / 2}=(-1-0)-(0-1)=0
\end{aligned}
$$
\begin{aligned}
&\int_{-\pi / 2}^{\pi / 2} \sin x d x=\int_{-\pi / 2}^{0} \sin x d x+\int_{0}^{\pi / 2} \sin x d x \\
&=-[\cos x]_{-\pi / 2}^{0}-[\cos x]_{0}^{\pi / 2}=(-1-0)-(0-1)=0
\end{aligned}
$$
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