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Question Number 218709 by Nicholas666 last updated on 14/Apr/25

        ∫_0 ^(π/2)  (dθ/( (√(sinθ +cosθ))))

$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{{sin}\theta\:+{cos}\theta}} \\ $$$$ \\ $$

Answered by Nicholas666 last updated on 14/Apr/25

    ∫_0 ^(π/2)  (dθ/( (√(sinθ + cosθ))))  solution;   ∫_0 ^(π/2)  (dθ/( (√((√2)sin(θ+(π/4)))))) = (1/(^4 (√2))) ∫_0 ^(π/2)  (dθ/( (√(sin(θ+π/4)))))    (1/(^4 (√2))) ∫_(π/4) ^(3π/4)  (du/( (√(sin u)))) = (1/(^4 (√2))) × 2.62206    ((2.62206)/(1.1892)) = 2.2049

$$\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{{sin}\theta\:+\:{cos}\theta}} \\ $$$${solution}; \\ $$$$\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{\sqrt{\mathrm{2}}{sin}\left(\theta+\frac{\pi}{\mathrm{4}}\right)}}\:=\:\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{{sin}\left(\theta+\pi/\mathrm{4}\right)}}\:\: \\ $$$$\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{2}}}\:\int_{\pi/\mathrm{4}} ^{\mathrm{3}\pi/\mathrm{4}} \:\frac{{du}}{\:\sqrt{{sin}\:{u}}}\:=\:\frac{\mathrm{1}}{\:^{\mathrm{4}} \sqrt{\mathrm{2}}}\:×\:\mathrm{2}.\mathrm{62206}\: \\ $$$$\:\frac{\mathrm{2}.\mathrm{62206}}{\mathrm{1}.\mathrm{1892}}\:=\:\mathrm{2}.\mathrm{2049} \\ $$$$\: \\ $$

Answered by Ghisom last updated on 14/Apr/25

∫_0 ^(π/2) (dθ/( (√(cos θ +sin θ))))=2∫_(π/4) ^(π/2) (dθ/( (√(cos θ +sin θ))))=       [t=(x/2)−(π/8)]  =2^(7/4) ∫_0 ^(π/8) (dt/( (√(1−2sin^2  t))))=2^(7/4) [F (t∣2)]_0 ^(π/8) =  =2^(7/4) F ((π/8)∣2) ≈1.39739529927

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta}}=\mathrm{2}\underset{\pi/\mathrm{4}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{d}\theta}{\:\sqrt{\mathrm{cos}\:\theta\:+\mathrm{sin}\:\theta}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}\right] \\ $$$$=\mathrm{2}^{\mathrm{7}/\mathrm{4}} \underset{\mathrm{0}} {\overset{\pi/\mathrm{8}} {\int}}\frac{{dt}}{\:\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{t}}}=\mathrm{2}^{\mathrm{7}/\mathrm{4}} \left[\mathrm{F}\:\left({t}\mid\mathrm{2}\right)\right]_{\mathrm{0}} ^{\pi/\mathrm{8}} = \\ $$$$=\mathrm{2}^{\mathrm{7}/\mathrm{4}} \mathrm{F}\:\left(\frac{\pi}{\mathrm{8}}\mid\mathrm{2}\right)\:\approx\mathrm{1}.\mathrm{39739529927} \\ $$

Commented by Nicholas666 last updated on 16/Apr/25

thank you for your solution

$${thank}\:{you}\:{for}\:{your}\:{solution} \\ $$

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