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Question Number 27400 by Tinkutara last updated on 06/Jan/18

$$Evaluate\underset{r}{\stackrel{0}{\int }}\sqrt{\frac{x}{r-x}}dx$$

Commented by Tinkutara last updated on 07/Jan/18

$$ButthisintegralwasusedinaPhysics$$$$questionwhereitsvalue{can}^{\prime }tbe$$$$negative.Itshouldbeonly\frac{\pi r}{2}.Itis$$$$usedinquestion27445.$$

Commented by abdo imad last updated on 06/Jan/18

$$letdothech.\sqrt{\frac{x}{r-x}}=t\iff \frac{x}{r-x}=t^2\iff x=(r-x)t^2$$$$\iff x+{xt}^2={rt}^2\iff (1+t^2)x={rt}^2\iff x=\frac{{rt}^2}{1+t^2}$$$$x=\frac{r(1+t^2-1)}{1+t^2}=r-\frac{r}{1+t^2}\iff \frac{dx}{dt}=\frac{2rt}{{(1+t^2)}^2}$$$${\int }_r^0\sqrt{\frac{x}{r-x}}dx={\int }_{\propto }^{0}t\frac{2rt}{{(1+t^2)}^2}dt=-2r{\int }_0^{\mathrm{\infty }}\frac{t^2}{{(1+t^2)}^2}dt$$$$but{\int }_0^{\mathrm{\infty }}\frac{t^2}{{(1+t^2)}^2}dt={\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}-{\int }_0^{\mathrm{\infty }}\frac{dt}{{(1+t^2)}^2}$$$$=\frac{\pi }{2}-{\int }_0^{\mathrm{\infty }}\frac{dt}{{(1+t^2)}^2}andbyresidustheorem$$$${\int }_0^{\mathrm{\infty }}\frac{dt}{{(1+t^2)}^2}=\frac{1}{2}{\int }_R\frac{dt}{{(1+t^2)}^2}=\frac{1}{2}2i\pi Res(f,i)with$$$$f\left(z\right)=\frac{1}{{(1+z^2)}^2}=\frac{1}{{(z-i)}^2{(z+i)}^2}andiisadoublepoleoff$$$$Res(f,i)={lim}_{z->i}\frac{1}{(2-1)!}{\left({(z-i)}^{2}f\left(z\right)\right)}^{,}$$$$={lim}_{z->i}{\left({(z+i)}^{-2}\right)}^{,}={lim}_{z->i}-2{(z+i)}^{-3}$$$$=-2{\left(2i\right)}^{-3}=-2\frac{1}{{\left(2i\right)}^3}=\frac{1}{4i}$$$${\int }_0^{\mathrm{\infty }}\frac{dt}{{(1+t^2)}^2}=i\pi \frac{1}{4i}=\frac{\pi }{4}$$$${\int }_r^0\sqrt{\frac{x}{r-x}}dx=-2r(\frac{\pi }{2}-\frac{\pi }{4})=-2r\frac{\pi }{4}=\frac{-r\pi }{2}.$$

Commented by mrW1 last updated on 07/Jan/18

$$theresultaboveiscorrect.$$$$$$$${\int }_r^0\sqrt{...}dxisnegative$$$${\int }_0^r\sqrt{...}dxispositive$$

Commented by abdo imad last updated on 07/Jan/18

$$wemusthave0<x<rbecauseof\sqrt{...}\Rightarrow {\int }_r^o(...)dx=-{\int }_0^r(...)⩽0$$

Answered by prakash jain last updated on 07/Jan/18

$$x=r{\cos }^2\theta$$$$dx=-2r\cos \theta \sin \theta d\theta$$$$\int -2r\cos \theta \sin \theta .\frac{\cos \theta }{\sin \theta }d\theta$$$$=-r\int {2\cos }^2\theta d\theta$$$$=-r\int (1+\cos 2\theta )d\theta$$$$=-r[\theta +\frac{\sin 2\theta }{2}]+C$$$$=-r[{\cos }^{-1}\sqrt{\frac{x}{r}}+\sqrt{\frac{x}{r}\times \frac{r-x}{r}}]+C$$$$=-r[{\cos }^{-1}\sqrt{\frac{x}{r}}+\frac{1}{r}\sqrt{x(r-x)}]+C$$$$takinglimits$$$$=-r[\frac{\pi }{2}-0]=-\frac{\pi r}{2}$$

Commented by mrW1 last updated on 07/Jan/18

$$veryniceway!$$

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