Question

$\int \sqrt{\frac{x}{1-x}}dx$ is equal to
(a) ${\sin }^{-1}\sqrt{x}+C$

(b) ${\sin }^{-1}\left\{\sqrt{x}-\sqrt{x\left(1-x\right)}\right\}+C$

(c) ${\sin }^{-1}\left\{\sqrt{x\left(1-x\right)}\right\}+C$

(d) ${\sin }^{-1}\sqrt{x}-\sqrt{x\left(1-x\right)}+C$

Answer 1

(d) ${\sin }^{-1}\sqrt{x}-\sqrt{x\left(1-x\right)}+C$

$$\begin{gathered} \mathrm{Let}I=\int \sqrt{\frac{x}{1-x}}dx \\ \mathrm{Putting}\sqrt{x}=\sin \theta \\ \Rightarrow x={\sin }^2\theta \\ \Rightarrow dx=2\sin \theta \cos \theta d\theta \\ \Rightarrow dx=\sin \left(2\theta \right)d\theta \\ \therefore I=\int \sqrt{\frac{{\sin }^2\theta }{1-{\sin }^2\theta }}\times \sin \left(2\theta \right)·d\theta \\ =\int \frac{\sin \theta }{\cos \theta }\times 2\sin \theta ·\cos \theta d\theta \\ =\int 2{\sin }^2\theta ·d\theta \\ =\int \left(1-\cos 2\theta \right)d\theta \\ =\theta -\frac{\sin \left(2\theta \right)}{2}+C \\ =\theta -\frac{2\sin \theta \cos \theta }{2}+C\mathit{} \\ =\theta -\sin \theta \sqrt{1-{\sin }^2\theta }+C \\ ={\sin }^{-1}\sqrt{x}-\sqrt{x}\sqrt{1-x}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\mathit{\because }\mathit{}\theta ={\sin }^{\mathit{-}\mathit{1}}\sqrt{x}\right) \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}{\sin }^{-1}\mathit{}\sqrt{x}\mathit{-}\sqrt{x\left(1\mathit{-}x\right)}+C\mathit{}\mathit{}\mathit{} \end{gathered}$$