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$\int_{-\pi / 2}^{\pi / 2}(2 \sin |x|+\cos |x|) d x=$
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Verified Answer
The correct answer is:
6
$$
\begin{aligned}
& \text { (b) } I=\int_{-\pi / 2}^{\pi / 2}(2 \sin |x|+\cos |x|) d x \\
& =2 \int_0^{\pi / 2}(2 \sin x+\cos x) d x \\
& =2[-2 \cos x+\sin x]_0^{\pi / 2} \\
& =2[(-2 \times 0)+1-(-2 \times 1)+0] \\
& =2[1+2]=6
\end{aligned}
$$
\begin{aligned}
& \text { (b) } I=\int_{-\pi / 2}^{\pi / 2}(2 \sin |x|+\cos |x|) d x \\
& =2 \int_0^{\pi / 2}(2 \sin x+\cos x) d x \\
& =2[-2 \cos x+\sin x]_0^{\pi / 2} \\
& =2[(-2 \times 0)+1-(-2 \times 1)+0] \\
& =2[1+2]=6
\end{aligned}
$$
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