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

Question

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { (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 	imes 0)+1-(-2 	imes 1)+0] \ & =2[1+2]=6 \end{aligned} $

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

Solution