Question Number 67572 by aliesam last updated on 28/Aug/19
$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left\{{sin}\mid{x}\mid+{cos}\mid{x}\mid\right\}\:{dx} \\ $$
Commented by mathmax by abdo last updated on 28/Aug/19
![∫_(−(π/2)) ^(π/2) (sin∣x∣ +cos∣x∣)dx =2 ∫_0 ^(π/2) {sinx +cosx} dx =2[−cosx +sinx]_0 ^(π/2) =2{1−(−1)} =4](Q67580.png)
$$\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\left({sin}\mid{x}\mid\:+{cos}\mid{x}\mid\right){dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left\{{sinx}\:+{cosx}\right\}\:{dx} \\ $$$$=\mathrm{2}\left[−{cosx}\:+{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\mathrm{2}\left\{\mathrm{1}−\left(−\mathrm{1}\right)\right\}\:=\mathrm{4} \\ $$