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Question Number 132925 by metamorfose last updated on 17/Feb/21

(∫_0 ^(π/2) (√(sin(x)))dx)^2 +(∫_0 ^(π/2) (√(cos(x)))dx)^2 <8((√2)−1)

$$\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}{dx}\right)^{\mathrm{2}} +\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}\left({x}\right)}{dx}\right)^{\mathrm{2}} <\mathrm{8}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$

Answered by Dwaipayan Shikari last updated on 17/Feb/21

Φ=∫_0 ^(π/2) (√(sin(x))) dx=2((Γ((3/4))Γ((1/2)))/(Γ((1/4))))=2(√2)π(1/(Γ^2 ((1/4))))  Ψ=∫_0 ^(π/2) (√(cos(x)))dx=∫_0 ^(π/2) (√(sin(x)))dx=2(√2)π.(1/(Γ^2 ((1/4))))  Φ^2 +Ψ^2 =((16π^2 )/(Γ^4 ((1/4))))

$$\Phi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}\:{dx}=\mathrm{2}\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}=\mathrm{2}\sqrt{\mathrm{2}}\pi\frac{\mathrm{1}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$\Psi=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{cos}\left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{sin}\left({x}\right)}{dx}=\mathrm{2}\sqrt{\mathrm{2}}\pi.\frac{\mathrm{1}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$\Phi^{\mathrm{2}} +\Psi^{\mathrm{2}} =\frac{\mathrm{16}\pi^{\mathrm{2}} }{\Gamma^{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$

Commented by Ñï= last updated on 17/Feb/21

Γ((1/2))=(√π)....?  Γ(1−x)Γ(x)=(π/(sinπx))...?  Confused why you got that result.

$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}....? \\ $$$$\Gamma\left(\mathrm{1}−{x}\right)\Gamma\left({x}\right)=\frac{\pi}{{sin}\pi{x}}...? \\ $$$${Confused}\:{why}\:{you}\:{got}\:{that}\:{result}. \\ $$

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