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