Question:
Mark the correct alternative in each of the following:
$\int \sqrt{\frac{x}{1-x}} d x$ is equal to
A. $\sin ^{-1} \sqrt{\mathrm{x}}+\mathrm{C}$
B. $\sin ^{-1}(\sqrt{x}-\sqrt{x(1-x)})+C$
C. $\sin ^{-1}\{\sqrt{x(1-x)}\}+C$
D. $\sin ^{-1} \sqrt{x}-\sqrt{x(1-x)}+C$
Solution:
let $x=(\sin t)^{2} ;(d x=2 \sin t \cos t d t)$
$I=\int \sqrt{\frac{(\sin t)^{2}}{1-(\sin t)^{2}}} \times 2 \sin t \cos t d t$
$\mathrm{I}=\int(\sin \mathrm{t})^{2} \mathrm{dt}$
$\mathrm{I}=\int(1-\cos 2 \mathrm{t}) \mathrm{dt}$
$\mathrm{I}=\int 1 \mathrm{dt}-\int \cos 2 \mathrm{t} \mathrm{dt}$
$I=t-\frac{\sin 2 t}{2}+c\left[\mathrm{t}=\sin ^{-1} \sqrt{\mathrm{x}}\right](\cos \mathrm{t}=\sqrt{1-\mathrm{x}})$
$I=\sin ^{-1}(\sqrt{x})-(\sqrt{x} \sqrt{1-x})+c$