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Question Number 137379 by oooooooo last updated on 02/Apr/21

Answered by Ar Brandon last updated on 02/Apr/21

$$\mathrm{I}={\int }_0^{\frac{\mathrm{a}}{2}}\sqrt{\frac{\mathrm{x}}{\mathrm{a}-\mathrm{x}}}\mathrm{dx}={\int }_0^{\frac{\mathrm{a}}{2}}\frac{{\mathrm{x}}^{\frac{1}{2}}}{\sqrt{\mathrm{a}-\mathrm{x}}}\mathrm{dx}$$$$\mathrm{u}={\mathrm{x}}^{\frac{1}{2}}\Rightarrow \mathrm{du}=\frac{1}{2}{\mathrm{x}}^{-\frac{1}{2}}\mathrm{dx}$$$$\mathrm{I}=2{\int }_0^{\sqrt{\frac{\mathrm{a}}{2}}}\frac{{\mathrm{u}}^2}{\sqrt{\mathrm{a}-{\mathrm{u}}^2}}\mathrm{du}=2{\int }_0^{\sqrt{\frac{\mathrm{a}}{2}}}\{\frac{\mathrm{a}}{\sqrt{\mathrm{a}-{\mathrm{u}}^2}}-\sqrt{\mathrm{a}-{\mathrm{u}}^2}\}\mathrm{du}$$$$\mathrm{Let}\mathrm{u}=\sqrt{\mathrm{a}}\sin \theta ...$$

Commented by mathmax by abdo last updated on 02/Apr/21

$$\mathrm{\Phi }={\int }_0^{\frac{\mathrm{a}}{2}}\sqrt{\frac{\mathrm{x}}{\mathrm{a}-\mathrm{x}}}\mathrm{dx}\mathrm{changement}\sqrt{\frac{\mathrm{x}}{\mathrm{a}-\mathrm{x}}}=\mathrm{t}\mathrm{give}\frac{\mathrm{x}}{\mathrm{a}-\mathrm{x}}={\mathrm{t}}^2\Rightarrow$$$$\mathrm{x}={\mathrm{at}}^2-{\mathrm{t}}^2\mathrm{x}\Rightarrow (1+{\mathrm{t}}^2)\mathrm{x}={\mathrm{at}}^2\Rightarrow \mathrm{x}=\frac{{\mathrm{at}}^2}{1+{\mathrm{t}}^2}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2\mathrm{at}(1+{\mathrm{t}}^2)-{\mathrm{at}}^2\left(2\mathrm{t}\right)}{{(1+{\mathrm{t}}^2)}^2}$$$$=\frac{2\mathrm{at}}{{(1+{\mathrm{t}}^2)}^2}\Rightarrow \mathrm{\Phi }={\int }_0^1\mathrm{t}\times \frac{2\mathrm{at}}{{(1+{\mathrm{t}}^2)}^2}\mathrm{dt}$$$$=2\mathrm{a}{\int }_0^1\frac{{\mathrm{t}}^2}{{(1+{\mathrm{t}}^2)}^2}\mathrm{dt}=2\mathrm{a}{\int }_0^1\frac{1+{\mathrm{t}}^2-1}{{(1+{\mathrm{t}}^2)}^2}\mathrm{dt}$$$$=2\mathrm{a}{\int }_0^1\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}-2\mathrm{a}{\int }_0^1\frac{\mathrm{dt}}{{(1+{\mathrm{t}}^2)}^2}\mathrm{we}\mathrm{have}$$$${\int }_0^1\frac{\mathrm{dt}}{1+{\mathrm{t}}^2}={\left[\mathrm{arctant}\right]}_0^1=\frac{\pi }{4}$$$${\int }_0^1\frac{\mathrm{dt}}{{(1+{\mathrm{t}}^2)}^2}{=}_{\mathrm{t}=\mathrm{tanu}}{\int }_0^{\frac{\pi }{4}}\frac{1+{\tan }^2\mathrm{u}}{{(1+{\tan }^2\mathrm{u})}^2}\mathrm{du}={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{du}}{1+{\tan }^2\mathrm{u}}$$$$={\int }_0^{\frac{\pi }{4}}{\cos }^2\mathrm{u}\mathrm{du}=\frac{1}{2}{\int }_0^{\frac{\pi }{4}}(1+\cos \left(2\mathrm{u}\right))\mathrm{du}=\frac{\pi }{8}+\frac{1}{4}{\left[\sin \left(2\mathrm{u}\right)\right]}_0^{\frac{\pi }{4}}$$$$=\frac{\pi }{8}+\frac{1}{4}\Rightarrow \mathrm{\Phi }=\frac{\pi \mathrm{a}}{2}-2\mathrm{a}(\frac{\pi }{8}+\frac{1}{4})=\frac{\pi \mathrm{a}}{2}-\frac{\pi \mathrm{a}}{4}+\frac{\mathrm{a}}{2}$$$$=\frac{\pi \mathrm{a}}{4}+\frac{\mathrm{a}}{2}=(\frac{1}{2}+\frac{\pi }{4})\mathrm{a}$$$$$$

Answered by Dwaipayan Shikari last updated on 03/Apr/21

$$x=\frac{a}{2}u\Rightarrow 1=\frac{a}{2}.\frac{du}{dx}$$$$=\frac{a}{2}{\int }_0^1\sqrt{\frac{\frac{a}{2}u}{a-\frac{a}{2}u}}du$$$$=\frac{a}{2}{\int }_0^1\sqrt{\frac{u}{2-u}}du=\frac{a}{2}{\int }_0^1\sqrt{\frac{1-u}{1+u}}du$$$$=\frac{a}{2}{\int }_0^1\frac{1}{\sqrt{1-u^2}}+\frac{a}{2}{\int }_0^1\frac{-2u}{\sqrt{1-u^2}}du$$$$=\frac{a}{2}\left(\frac{\pi }{2}\right)+\frac{a}{2}{\left(\sqrt{1-u^2}\right)}_0^1=\frac{\pi -2}{4}a$$

Answered by MJS_new last updated on 03/Apr/21

$$\int \sqrt{\frac{x}{a-x}}dx=$$$$[t=\sqrt{\frac{x}{a-x}}\to dx=\frac{2at}{{(t^2+1)}^2}dt]$$$$=2a\int \frac{t^2}{{(t^2+1)}^2}dt=2a\int \frac{dt}{t^2+1}-2a\int \frac{dt}{(t^2+1)}=$$$$=a\arctan t-\frac{at}{t^2+1}=$$$$=a\arctan \sqrt{\frac{x}{a-x}}-\sqrt{x(a-x)}+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{(\pi -2)a}{4}$$

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