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Question Number 118566 by benjo_mathlover last updated on 18/Oct/20

$$solve\underset{0}{\stackrel{1}{\int }}\frac{dx}{\sqrt{x}\sqrt{1-x}}.$$

Commented by benjo_mathlover last updated on 18/Oct/20

$$inotherwaybysubstitutetrigonometri$$$$lettingx=\sin {}^2\theta \to dx=2\sin \theta \cos \theta d\theta$$$$forx=0\to \theta =0,x=1\to \theta =\frac{\pi }{2}$$$$then\sqrt{x}=\mid \sin \theta \mid =\sin \theta and\sqrt{1-x}=\mid \cos \theta \mid =\cos \theta$$$$sotheintegralbecomes$$$$\int {}_0^{\pi /2}\frac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }d\theta =2\int {}_0^{\pi /2}d\theta =2\theta {\mid }_0^{\frac{\pi }{2}}=\pi$$

Commented by Dwaipayan Shikari last updated on 18/Oct/20

$${\int }_0^1{\left(x\right)}^{-\frac{1}{2}}{(1-x)}^{-\frac{1}{2}}dx$$$$={\int }_0^1{\left(x\right)}^{\frac{1}{2}-1}{(1-x)}^{\frac{1}{2}-1}dx$$$$=\beta (\frac{1}{2},\frac{1}{2})=\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right).\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(1\right)}=\frac{\sqrt{\pi }.\sqrt{\pi }}{1}=\pi ⊛$$$$Anotherway★$$

Answered by TANMAY PANACEA last updated on 18/Oct/20

$$t^2=1-x$$$${\int }_1^0\frac{-2tdt}{\sqrt{1-t^2}\times t}$$$$2{\int }_0^1\frac{dt}{\sqrt{1-t^2}}\to 2\mid {sin}^{-1}t{\mid }_0^1=2\times \frac{\pi }{2}=\pi$$

Answered by Dwaipayan Shikari last updated on 18/Oct/20

$${\int }_0^1\frac{dx}{\sqrt{x}.\sqrt{1-x}}1-x=t^2\Rightarrow -1=2t\frac{dt}{dx}$$$$-{\int }_1^0\frac{2tdt}{t\sqrt{1-t^2}}=2{\int }_0^1\frac{1}{\sqrt{1-t^2}}dt$$$$=\pi$$

Answered by MJS_new last updated on 18/Oct/20

$$\underset{0}{\stackrel{1}{\int }}\frac{dx}{\sqrt{x}\sqrt{1-x}}=$$$$[t=\frac{\sqrt{1-x}}{\sqrt{x}}\to dx=-2\sqrt{x^3}\sqrt{1-x}dt]$$$$=-2\underset{+\mathrm{\infty }}{\stackrel{0}{\int }}\frac{dt}{t^2+1}=2\underset{0}{\stackrel{+\mathrm{\infty }}{\int }}\frac{dt}{t^2+1}=$$$$=2{\left[\arctan t\right]}_0^{+\mathrm{\infty }}=\pi$$

Answered by 1549442205PVT last updated on 18/Oct/20

$$\mathrm{Put}\frac{1}{\mathrm{x}}-1={\mathrm{t}}^2\Rightarrow 2\mathrm{tdt}=-\frac{1}{{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{t}=\frac{\sqrt{1-\mathrm{x}}}{\sqrt{\mathrm{x}}},\mathrm{dx}=-\frac{2\mathrm{t}}{{({\mathrm{t}}^2+1)}^2}\mathrm{dt},\mathrm{x}=\frac{1}{{\mathrm{t}}^2+1}$$$$\underset{0}{\stackrel{1}{\int }}\frac{dx}{\sqrt{x}\sqrt{1-x}}=-{\int }_{\mathrm{\infty }}^0\frac{{\mathrm{t}}^2+1}{\mathrm{t}}\times \frac{2\mathrm{tdt}}{{({\mathrm{t}}^2+1)}^2}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dt}}{{\mathrm{t}}^2+1}={2\tan }^{-1}\left(\mathrm{t}\right)=\pi$$

Commented by MJS_new last updated on 18/Oct/20

$$\mathrm{is}\mathrm{it}\mathrm{just}\mathrm{coincidence}\mathrm{or}\mathrm{do}\mathrm{you}\mathrm{repost}\mathrm{my}$$$$\mathrm{answers}\mathrm{for}\mathrm{a}\mathrm{special}\mathrm{reason}?$$

Commented by MJS_new last updated on 18/Oct/20

$$\mathrm{sorry}\mathrm{I}\mathrm{asked}.\mathrm{you}\mathrm{posted}\mathrm{many}\mathrm{good}\mathrm{answers}.$$

Answered by Bird last updated on 18/Oct/20

$$I={\int }_0^1\frac{dx}{\sqrt{x}\sqrt{1-x}}wedothechang.$$$$\sqrt{x}=t\Rightarrow I={\int }_0^1\frac{2tdt}{t\sqrt{1-t^2}}$$$$=2{\int }_0^1\frac{dt}{\sqrt{1-t^2}}=2{\left[arcsint\right]}_0^1$$$$=2\times \frac{\pi }{2}=\pi$$

Answered by mindispower last updated on 18/Oct/20

$$=2{\int }_0^1.\frac{1}{2\sqrt{x}}.\frac{1}{\sqrt{1-{\left(\sqrt{x}\right)}^2}}=2{\left[arcsin\left(\sqrt{x}\right)\right]}_0^1=2.\frac{\pi }{2}=1\pi$$

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