Question Number 95227 by bobhans last updated on 24/May/20
$$\underset{−\pi} {\overset{\pi} {\int}}\:\mid\mathrm{sin}\:\mathrm{x}\:+\:\mathrm{cos}\:\mathrm{x}\:\mid\:\mathrm{dx}\:=?\: \\ $$
Answered by john santu last updated on 24/May/20
Commented by john santu last updated on 24/May/20
![∫_(−π) ^(−(π/4)) (sin x+cos x) dx + ∫_(−(π/4)) ^((3π)/4) (sin x+cos x) dx+∫_((3π)/4) ^π (sin x+cos x) dx = 2∫_(−(π/4)) ^((3π)/4) (sin x+cos x) dx = 2 [ −cos x+sin x ]_(−(π/4)) ^((3π)/4) = 2{(√2)−(−(√2))} = 4(√2)](https://www.tinkutara.com/question/Q95229.png)
$$\underset{−\pi} {\overset{−\frac{\pi}{\mathrm{4}}} {\int}}\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\:\mathrm{dx}\:+\:\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\frac{\mathrm{3}\pi}{\mathrm{4}}} {\int}}\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\:\mathrm{dx}+\underset{\frac{\mathrm{3}\pi}{\mathrm{4}}} {\overset{\pi} {\int}}\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\:\mathrm{dx}\:= \\ $$$$\mathrm{2}\underset{−\frac{\pi}{\mathrm{4}}} {\overset{\frac{\mathrm{3}\pi}{\mathrm{4}}} {\int}}\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\:\mathrm{dx}\:= \\ $$$$\mathrm{2}\:\left[\:−\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}\:\right]_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} = \\ $$$$\mathrm{2}\left\{\sqrt{\mathrm{2}}−\left(−\sqrt{\mathrm{2}}\right)\right\}\:=\:\mathrm{4}\sqrt{\mathrm{2}}\: \\ $$