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Question Number 92668 by john santu last updated on 08/May/20

$$\underset{0}{\stackrel{\mathrm{p}}{\int }}\sqrt{\frac{\mathrm{x}}{\mathrm{c}-\mathrm{x}}}\mathrm{dx}?$$

Commented by john santu last updated on 08/May/20

$$\mathrm{set}\mathrm{t}=\sqrt{\frac{x}{c-x}},x=\frac{{ct}^2}{1+t^2}$$$$dx=\frac{2ctdt}{{(1+t^2)}^2}$$$$I=\underset{0}{\stackrel{\sqrt{\frac{\mathrm{p}}{\mathrm{c}-\mathrm{p}}}}{\int }}\frac{2{ct}^{2}dt}{{(1+t^2)}^2}$$$$I=-\mathrm{c}{\left[\frac{\mathrm{t}}{1+{\mathrm{t}}^2}\right]}_0^{\sqrt{\frac{\mathrm{p}}{\mathrm{c}-\mathrm{p}}}}+\mathrm{c}{\left[{\tan }^{-1}\mathrm{t}\right]}_0^{\sqrt{\frac{\mathrm{p}}{\mathrm{c}-\mathrm{p}}}}$$$$=-\mathrm{c}\left[\frac{\sqrt{\frac{\mathrm{p}}{\mathrm{c}-\mathrm{p}}}}{1+\frac{\mathrm{p}}{\mathrm{c}-\mathrm{p}}}\right]+\mathrm{c}\left[{\tan }^{-1}\sqrt{\frac{\mathrm{p}}{\mathrm{c}-\mathrm{p}}}\right]$$

Commented by mathmax by abdo last updated on 08/May/20

$$I={\int }_0^p\sqrt{\frac{x}{c-x}}dxchangementx={csin}^{2}tgive$$$$I={\int }_0^{arcsin\left(\sqrt{\frac{p}{c}}\right)}\sqrt{\frac{{csin}^{2}t}{{ccos}^{2}t}}2csintcost$$$$=2c{\int }_0^{arcsin\left(\sqrt{\frac{p}{c}}\right)}\frac{sint}{cost}sintcostdt=2c{\int }_0^{arcsin\left(\sqrt{\frac{p}{c}}\right)}\frac{1-cos\left(2t\right)}{2}dt$$$$=carcsin\left(\sqrt{\frac{p}{c}}\right)-\frac{c}{2}{\left[sin\left(2t\right)\right]}_0^{arcsin\left(\sqrt{\frac{p}{c}}\right)}+\lambda$$$$=carcsin\left(\sqrt{\frac{p}{c}}\right)-c{\left[sint\sqrt{1-{sin}^{2}t}\right]}_0^{arcsin\left(\sqrt{\frac{p}{c}}\right)}+\lambda$$$$=carcsin\left(\sqrt{\frac{p}{c}}\right)-c\sqrt{\frac{p}{c}}\sqrt{1-\frac{p}{c}}+\lambda$$$$=carcsin\left(\sqrt{\frac{p}{c}}\right)-\sqrt{p(c-p)}+\lambda$$

Answered by behi83417@gmail.com last updated on 08/May/20

$$\mathrm{x}=\mathrm{c}.{\cos }^2\mathrm{t}\Rightarrow \mathrm{dx}=-2\mathrm{csint}.\mathrm{costdt}$$$$\Rightarrow \mathrm{I}=\int \frac{\mathrm{cost}}{\mathrm{sint}}.(-2\mathrm{c}.\mathrm{sint}.\mathrm{cost})\mathrm{dt}=$$$$=-\int {2\mathrm{ccos}}^2\mathrm{tdt}=-\mathrm{c}\int (1+\cos 2\mathrm{t})\mathrm{dt}=$$$$=-\mathrm{c}(\mathrm{t}+\mathrm{sint}.\mathrm{cost})+\mathrm{const}.=$$$$=-\mathrm{c}.{\cos }^{-1}\sqrt{\frac{\mathrm{x}}{\mathrm{c}}}-\sqrt{\mathrm{cx}-{\mathrm{x}}^2}+\mathrm{const}.$$$$[\mathrm{I}=\underset{0}{\stackrel{\mathrm{p}}{\int }}\sqrt{\frac{\mathrm{x}}{\mathrm{c}-\mathrm{x}}}=\frac{\mathbf{c}.\pi }{2}-\mathbf{c}.{\mathbf{cos}}^{-1}\sqrt{\frac{\mathbf{p}}{\mathbf{c}}}-\sqrt{\mathbf{c}.\mathbf{p}-{\mathbf{p}}^2}]$$$$[\mathrm{cost}=\sqrt{\frac{\mathrm{x}}{\mathrm{c}}},\mathrm{sint}=\sqrt{1-\frac{\mathrm{x}}{\mathrm{c}}}]$$

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