Question Number 218709 by Nicholas666 last updated on 14/Apr/25 | ||
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$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{{sin}\theta\:+{cos}\theta}} \\ $$$$ \\ $$ | ||
Answered by Nicholas666 last updated on 14/Apr/25 | ||
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$$\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{{sin}\theta\:+\:{cos}\theta}} \\ $$$${solution}; \\ $$$$\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{\sqrt{\mathrm{2}}{sin}\left(\theta+\frac{\pi}{\mathrm{4}}\right)}}\:=\:\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{{sin}\left(\theta+\pi/\mathrm{4}\right)}}\:\: \\ $$$$\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{2}}}\:\int_{\pi/\mathrm{4}} ^{\mathrm{3}\pi/\mathrm{4}} \:\frac{{du}}{\:\sqrt{{sin}\:{u}}}\:=\:\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{2}}}\:×\:\mathrm{2}.\mathrm{62206}\: \\ $$$$\:\frac{\mathrm{2}.\mathrm{62206}}{\mathrm{1}.\mathrm{1892}}\:=\:\mathrm{2}.\mathrm{2049} \\ $$$$\: \\ $$ | ||
Answered by Ghisom last updated on 14/Apr/25 | ||
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$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta}}=\mathrm{2}\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}\right] \\ $$$$=\mathrm{2}^{\mathrm{7}/\mathrm{4}} \underset{\mathrm{0}} {\overset{\pi/\mathrm{8}} {\int}}\frac{{dt}}{\:\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{t}}}=\mathrm{2}^{\mathrm{7}/\mathrm{4}} \left[\mathrm{F}\:\left({t}\mid\mathrm{2}\right)\right]_{\mathrm{0}} ^{\pi/\mathrm{8}} = \\ $$$$=\mathrm{2}^{\mathrm{7}/\mathrm{4}} \mathrm{F}\:\left(\frac{\pi}{\mathrm{8}}\mid\mathrm{2}\right)\:\approx\mathrm{1}.\mathrm{39739529927} \\ $$ | ||
Commented by Nicholas666 last updated on 16/Apr/25 | ||
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$${thank}\:{you}\:{for}\:{your}\:{solution} \\ $$ | ||