Question Number 152088 by rexford last updated on 25/Aug/21 | ||
$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \mid{sinx}−{cosx}\mid \\ $$$${please}\:{help}\:{me}\:{out} \\ $$ | ||
Answered by Olaf_Thorendsen last updated on 25/Aug/21 | ||
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mid\mathrm{sin}{x}−\mathrm{cos}{x}\mid\:{dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mid\sqrt{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\mid\:{dx} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\:{dx} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\left[\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$\mathrm{I}\:=\:\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:=\:\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$ | ||
Answered by puissant last updated on 25/Aug/21 | ||
$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left({cosx}−{sinx}\right){dx}+\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \left({sinx}−{cosx}\right){dx} \\ $$$$=\left[{sinx}+{cosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\left[−{cosx}−{sinx}\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\:\sqrt{\mathrm{2}}−\mathrm{1}+\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$$ \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2} \\ $$$$ \\ $$$$=\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$ | ||
Commented by peter frank last updated on 25/Aug/21 | ||
$$\mathrm{good} \\ $$ | ||
Commented by rexford last updated on 26/Aug/21 | ||
$${thank}\:{you} \\ $$ | ||