Integration Questions

Question Number 38516 by rahul 19 last updated on 26/Jun/18

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mid\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\mid\mathrm{d}{x} \\$$

Commented bymath khazana by abdo last updated on 27/Jun/18

$${we}\:{have}\:{cosx}\:−{sinx}=\sqrt{\mathrm{2}}{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)\:\Rightarrow \\$$ $${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{2}}\mid{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)\mid{dx}\:\:{changement} \\$$ $${x}+\frac{\pi}{\mathrm{4}}={t}\:{give}\:\:{I}\:=\sqrt{\mathrm{2}}\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\mid{cost}\mid\:{dt} \\$$ $$=\sqrt{\mathrm{2}}\left\{\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:{cost}\:{dt}\:+\:\int_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:−{cost}\:{dt}\right\} \\$$ $$=\sqrt{\mathrm{2}}\left\{\:\:\left[{sint}\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \:\:−\left[{sint}\right]_{\frac{\pi}{\mathrm{2}}} ^{\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\right\} \\$$ $$=\sqrt{\mathrm{2}}\:\left\{\:\:\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\right)\right\} \\$$ $$=\sqrt{\mathrm{2}}\left\{\:\mathrm{2}−\sqrt{\mathrm{2}}\right)\:=\:\mathrm{2}\sqrt{\mathrm{2}}\:−\mathrm{2}\:= \\$$ $${I}\:=\mathrm{2}\sqrt{\mathrm{2}}\:−\mathrm{2} \\$$

Answered by MJS last updated on 26/Jun/18

$$\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x}=−\sqrt{\mathrm{2}}\mathrm{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right) \\$$ $$\mathrm{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{0}\:\Rightarrow\:{x}=\frac{\pi}{\mathrm{4}} \\$$ $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mid\mathrm{sin}\:{x}\:−\mathrm{cos}\:{x}\mid{dx}= \\$$ $$=\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right){dx}−\sqrt{\mathrm{2}}\underset{\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right){dx}= \\$$ $$=\mathrm{2}\sqrt{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{2}\sqrt{\mathrm{2}}\left[\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} = \\$$ $$=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2} \\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Jun/18

$$\mid{sinx}−{cosx}\mid \\$$ $$−\left({sinx}−{cosx}\right)\:\:\:\:\frac{\Pi}{\mathrm{4}}>{x}\geqslant\mathrm{0} \\$$ $$\:\left({sinx}−{cosx}\right)\:\:\:\:\:\frac{\Pi}{\mathrm{2}}>{x}\geqslant\frac{\Pi}{\mathrm{4}} \\$$ $${see}\:{the}\:{graph}\:{from}\: \\$$ $${value}\:{of}\:{cosx}>{sinx}\:\:\:{when}\:\:\frac{\Pi}{\mathrm{4}}>{x}\geqslant\mathrm{0} \\$$ $${value}\:{of}\:{sinx}>{cosx}\:{when}\:\:\frac{\Pi}{\mathrm{2}}>{x}\geqslant\frac{\Pi}{\mathrm{4}} \\$$ $$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \mid{sinx}−{cosx}\mid{dx} \\$$ $$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} −\left({sinx}−{cosx}\right){dx}+\int_{\frac{\Pi}{\mathrm{4}}} ^{\frac{\Pi}{\mathrm{2}}} \:\left({sinx}−{cosx}\right){dx} \\$$ $$\mid{cosx}+{sinx}\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{4}}} +\mid\left(−{cosx}−{sinx}\right)\mid_{\frac{\Pi}{\mathrm{4}}} ^{\frac{\Pi}{\mathrm{2}}} \\$$ $$\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}−\mathrm{1}−\mathrm{0}\right)−\left\{\left(\mathrm{0}+\mathrm{1}\right)−\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)\right\} \\$$ $$\frac{\mathrm{2}}{\sqrt{\mathrm{2}}}−\mathrm{1}−\mathrm{1}+\frac{\mathrm{2}}{\sqrt{\mathrm{2}}} \\$$ $$\mathrm{2}\sqrt{\mathrm{2}}\:−\mathrm{2} \\$$