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Question Number 125421 by bemath last updated on 11/Dec/20

 ∫_0 ^(100)  (dx/( (√(x(100−x))))) ?

$$\:\underset{\mathrm{0}} {\overset{\mathrm{100}} {\int}}\:\frac{{dx}}{\:\sqrt{{x}\left(\mathrm{100}−{x}\right)}}\:?\: \\ $$

Answered by liberty last updated on 11/Dec/20

I=∫_0 ^(100)  (dx/( (√x) (√(100−x)))) ; let (√x) = 10 sin t   dx = 100.(2sin t cos t) dt   I=∫_0 ^(π/2) ((200 sin t cos t)/(10 sin t (10 cos t))) dt = 2∫_0 ^(π/2) dt   I= 2×(π/2)=π

$${I}=\underset{\mathrm{0}} {\overset{\mathrm{100}} {\int}}\:\frac{{dx}}{\:\sqrt{{x}}\:\sqrt{\mathrm{100}−{x}}}\:;\:{let}\:\sqrt{{x}}\:=\:\mathrm{10}\:\mathrm{sin}\:{t}\: \\ $$$${dx}\:=\:\mathrm{100}.\left(\mathrm{2sin}\:{t}\:\mathrm{cos}\:{t}\right)\:{dt}\: \\ $$$${I}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{200}\:\mathrm{sin}\:{t}\:\mathrm{cos}\:{t}}{\mathrm{10}\:\mathrm{sin}\:{t}\:\left(\mathrm{10}\:\mathrm{cos}\:{t}\right)}\:{dt}\:=\:\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}{dt}\: \\ $$$${I}=\:\mathrm{2}×\frac{\pi}{\mathrm{2}}=\pi \\ $$

Answered by Dwaipayan Shikari last updated on 11/Dec/20

∫_0 ^(100) (dx/( (√(x(100−x)))))            u=100x  =∫_0 ^1 (du/( (√u)((√(1−u)))))=β((1/2),(1/2))=Γ^2 ((1/2))=π

$$\int_{\mathrm{0}} ^{\mathrm{100}} \frac{{dx}}{\:\sqrt{{x}\left(\mathrm{100}−{x}\right)}}\:\:\:\:\:\:\:\:\:\:\:\:{u}=\mathrm{100}{x} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{du}}{\:\sqrt{{u}}\left(\sqrt{\mathrm{1}−{u}}\right)}=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right)=\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\pi \\ $$

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