int_{-pi / 2}^{pi / 2}(2 sin |x|+cos |x|) d x | vedclass

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(B) Let $I = \int_{-\pi / 2}^{\pi / 2} (2 \sin |x| + \cos |x|) dx$. Since $f(x) = 2 \sin |x| + \cos |x|$ is an even function because $f(-x) = 2 \sin |

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Let $I = \int_{-\pi / 2}^{\pi / 2} (2 \sin |x| + \cos |x|) dx$. Since $f(x) = 2 \sin |x| + \cos |x|$ is an even function because $f(-x) = 2 \sin |-x| + \cos |-x| = 2 \sin |x| + \cos |x| = f(x)$,we can use the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$. $I = 2 \int_{0}^{\pi / 2} (2 \sin x + \cos x) dx$. Integrating the terms,we get $I = 2 [-2 \cos x + \sin x]_{0}^{\pi / 2}$. Evaluating at the limits,$I = 2 [(-2 \cos(\pi / 2) + \sin(\pi / 2)) - (-2 \cos(0) + \sin(0))]$. $I = 2 [(-2 	imes 0 + 1) - (-2 	imes 1 + 0)]$. $I = 2 [1 - (-2)] = 2 [1 + 2] = 2 	imes 3 = 6$.

$\int_{-\pi / 2}^{\pi / 2}(2 \sin |x|+\cos |x|) d x=$

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