int_{-pi / 2}^{pi / 2} sin |x| , dx is equal | vedclass

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(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$. Since $f(x) = \sin |x|$ is an even function because $f(-x) = \sin |-x| = \sin |x| = f(x)$,we c

Vedclass · Accepted Answer

$2$

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(C) Let $I = \int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$. Since $f(x) = \sin |x|$ is an even function because $f(-x) = \sin |-x| = \sin |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$. Thus,$I = 2 \int_{0}^{\pi / 2} \sin |x| \, dx$. For $x \in [0, \pi / 2]$,$|x| = x$,so $I = 2 \int_{0}^{\pi / 2} \sin x \, dx$. Evaluating the integral,$I = 2 [-\cos x]_{0}^{\pi / 2}$. $I = 2 [-\cos(\pi / 2) - (-\cos(0))]$. $I = 2 [0 - (-1)] = 2(1) = 2$.

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

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