Question

Integration of dx/√{x(1-x)} =?

Answer 1

Dear student
$$\begin{gathered} \int \frac{\mathrm{dx}}{\sqrt{\mathrm{x}\left(1-\mathrm{x}\right)}} \\ =\int \frac{1}{\sqrt{\mathrm{x}}\sqrt{1-\mathrm{x}}}\mathrm{dx} \\ \mathrm{Put}\mathrm{u}=\sqrt{\mathrm{x}}\Rightarrow {\mathrm{u}}^2=\mathrm{x} \\ \frac{\mathrm{du}}{\mathrm{dx}}=\frac{1}{2\sqrt{\mathrm{x}}}\Rightarrow 2\sqrt{\mathrm{x}}\mathrm{du}=\mathrm{dx}\Rightarrow 2\mathrm{udu}=\mathrm{dx} \\ \mathrm{So},\int \frac{1}{\sqrt{\mathrm{x}}\sqrt{1-\mathrm{x}}}\mathrm{dx}=\int \frac{1}{\mathrm{u}\sqrt{1-{\mathrm{u}}^2}}2\mathrm{udu} \\ =2\int \frac{\mathrm{du}}{\sqrt{1-{\mathrm{u}}^2}} \\ =2{\sin }^{-1}\left(\mathrm{u}\right)+\mathrm{C} \\ =2{\sin }^{-1}\left(\sqrt{\mathrm{x}}\right)+\mathrm{C} \end{gathered}$$
Regards