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Question Number 83834 by Power last updated on 06/Mar/20

Commented by niroj last updated on 06/Mar/20

$$\int \frac{1}{\sqrt{\mathrm{x}(\mathrm{a}-\mathrm{x})}}\mathrm{dx}$$$$=\int \frac{1}{\sqrt{\mathrm{ax}-{\mathrm{x}}^2}}\mathrm{dx}=\int \frac{1}{\sqrt{-({\mathrm{x}}^2-2\mathrm{x}.\frac{\mathrm{a}}{2}+\frac{{\mathrm{a}}^2}{4}-\frac{{\mathrm{a}}^2}{4})}}\mathrm{dx}$$$$=\int \frac{1}{\sqrt{-[{(\mathrm{x}-\frac{\mathrm{a}}{2})}^2-\frac{{\mathrm{a}}^2}{4}]}}\mathrm{dx}$$$$=\int \frac{1}{\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^2-{(\mathrm{x}-\frac{\mathrm{a}}{2})}^2}}\mathrm{dx}$$$$={\sin }^{-1}\frac{(\mathrm{x}-\frac{\mathrm{a}}{2})}{\frac{\mathrm{a}}{2}}+\mathrm{C}={\sin }^{-1}\frac{2\mathrm{x}-\mathrm{a}}{\mathrm{a}}+\mathrm{C}//.$$$$$$

Commented by mathmax by abdo last updated on 06/Mar/20

$$I=\int \frac{dx}{\sqrt{x(a-x)}}vhangement\sqrt{x}=tgivex=t^2\Rightarrow$$$$I=\int \frac{2tdt}{t\sqrt{a-t^2}}=2\int \frac{dt}{\sqrt{a-t^2}}{=}_{t=\sqrt{a}sinu}2\int \frac{\sqrt{a}cosudu}{\sqrt{a}cosu}$$$$=2u+K=2arcsin\left(\frac{t}{\sqrt{a}}\right)+k=2arcsin\left(\frac{\sqrt{x}}{\sqrt{a}}\right)+K$$

Answered by TANMAY PANACEA last updated on 06/Mar/20

$$t^2=a-x$$$$\int \frac{-2tdt}{\sqrt{(a-t^2)\times t^2}}$$$$-2\int \frac{dt}{\sqrt{a-t^2}}$$$$t=\sqrt{a}\times sin\alpha$$$$-2\int \frac{\sqrt{a}\times cos\alpha d\alpha }{\sqrt{a}\times cos\alpha }=-2\times \alpha$$$$-2\times {sin}^{-1}\left(\frac{t}{\sqrt{a}}\right)$$$$-2\times {sin}^{-1}\left(\sqrt{\frac{a-x}{a}}\right)+c$$$$plscheck$$

Commented by Power last updated on 06/Mar/20

$$\mathrm{thanks}$$

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