Question

${\int }_0^{\pi /2}(|\sin x|+|\cos x|)dx$

Answer 1

In the interval $[0,\pi /2],|\sin x|=\sin x$ and $|\cos x|=\cos x$

${\int }_0^{\pi /2}(|\sin x|+|\cos x|)dx={\int }_0^{\pi /2}\sin xdx+{\int }_0^{\pi /2}\cos xdx=[-\cos x{]}_0^{\pi /2}+[\sin x{]}_0^{\pi /2}=[-0+1]+[1-0]=2$