Question Number 108382 by bobhans last updated on 16/Aug/20
$$\:\:\:\frac{\boldsymbol{{bobhans}}}{\iddots\ddots} \\ $$$$\:\underset{\pi/\mathrm{2}} {\overset{\pi} {\int}}\mid\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\:\mid\:{dx}\:? \\ $$$$ \\ $$
Commented by PRITHWISH SEN 2 last updated on 16/Aug/20
$$\mathrm{by}\:\mathrm{shifting}\:\mathrm{the}\:\mathrm{entire}\:\mathrm{function}\:\mathrm{by}\:\pi\:\mathrm{we}\:\mathrm{get} \\ $$$$\int_{\frac{\mathrm{3}\pi}{\mathrm{2}}} ^{\mathrm{2}\pi} \left(\mathrm{cos}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\right)\mathrm{dx}=\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\mid_{\frac{\mathrm{3}\pi}{\mathrm{2}}} ^{\mathrm{2}\pi} \:=\:\mathrm{2} \\ $$
Answered by bemath last updated on 16/Aug/20
![((△((Be)/(Math))△)/△) we know that cos x−sin x < 0 for (π/2) < x < π then ∫_(π/2) ^π ∣cos x−sin x∣ dx = ∫_(π/2) ^π (sin x−cos x) dx = (−cos x−sin x)]_(π/2) ^π = 1−(−1) = 2](https://www.tinkutara.com/question/Q108383.png)
$$\:\:\:\:\frac{\bigtriangleup\frac{{Be}}{\mathcal{M}{ath}}\bigtriangleup}{\bigtriangleup} \\ $$$$\:{we}\:{know}\:{that}\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\:<\:\mathrm{0}\: \\ $$$${for}\:\frac{\pi}{\mathrm{2}}\:<\:{x}\:<\:\pi \\ $$$$\:{then}\:\underset{\pi/\mathrm{2}} {\overset{\pi} {\int}}\mid\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\mid\:{dx}\:= \\ $$$$\left.\underset{\pi/\mathrm{2}} {\overset{\pi} {\int}}\left(\mathrm{sin}\:{x}−\mathrm{cos}\:{x}\right)\:{dx}\:=\:\left(−\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} \\ $$$$=\:\mathrm{1}−\left(−\mathrm{1}\right)\:=\:\mathrm{2}\: \\ $$
Answered by 1549442205PVT last updated on 17/Aug/20
![cosx−sinx=(√2) (((√2)/2)cosx−((√2)/2)sinx) =(√2)(sin(π/4)cosx−cos(π/4)sinx) =(√(2 ))sin((π/4)−x)=−(√2) .sin(x−(π/4))<0 for x∈[(π/2);π].Hence, I=∫_(π/2) ^π ∣ cos x−sin x ∣ dx = ∫_(π/2) ^π (sinx−cos x) dx =(−cosx−sinx)∣ _(π/2)^π =(1−0)−(0−1) =2 or I=(√2)∫sin(x−(π/4))dx=−(√2)cos(x−(π/4))∣_(π/2) ^π =−(√2) (cos((3π)/4)−cos(π/4))=−(√2)(−((√2)/2)−((√2)/2)) =−(√2)×(−(√2) )=2](https://www.tinkutara.com/question/Q108441.png)
$$\mathrm{cosx}−\mathrm{sinx}=\sqrt{\mathrm{2}}\:\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cosx}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sinx}\right) \\ $$$$=\sqrt{\mathrm{2}}\left(\mathrm{sin}\frac{\pi}{\mathrm{4}}\mathrm{cosx}−\mathrm{cos}\frac{\pi}{\mathrm{4}}\mathrm{sinx}\right) \\ $$$$=\sqrt{\mathrm{2}\:}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−\mathrm{x}\right)=−\sqrt{\mathrm{2}}\:.\mathrm{sin}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)<\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{x}\in\left[\frac{\pi}{\mathrm{2}};\pi\right].\mathrm{Hence},\:\mathrm{I}=\underset{\pi/\mathrm{2}} {\overset{\pi} {\int}}\mid\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\:\mid\:{dx} \\ $$$$=\:\underset{\pi/\mathrm{2}} {\overset{\pi} {\int}}\left(\mathrm{sinx}−\mathrm{cos}\:{x}\right)\:\:{dx}\: \\ $$$$=\left(−\mathrm{cosx}−\mathrm{sinx}\right)\mid\:_{\frac{\pi}{\mathrm{2}}} ^{\pi} =\left(\mathrm{1}−\mathrm{0}\right)−\left(\mathrm{0}−\mathrm{1}\right) \\ $$$$=\mathrm{2} \\ $$$$\mathrm{or}\:\mathrm{I}=\sqrt{\mathrm{2}}\int\mathrm{sin}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)\mathrm{dx}=−\sqrt{\mathrm{2}}\mathrm{cos}\left(\mathrm{x}−\frac{\pi}{\mathrm{4}}\right)\mid_{\frac{\pi}{\mathrm{2}}} ^{\pi} \\ $$$$=−\sqrt{\mathrm{2}}\:\left(\mathrm{cos}\frac{\mathrm{3}\pi}{\mathrm{4}}−\mathrm{cos}\frac{\pi}{\mathrm{4}}\right)=−\sqrt{\mathrm{2}}\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$=−\sqrt{\mathrm{2}}×\left(−\sqrt{\mathrm{2}}\:\right)=\mathrm{2} \\ $$