Question

The value of $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$ is
(where $C$ is constant of integration)

• A. $2\sqrt{1-x}-{\cos }^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}+C$
• B. $-2\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}+C$
• C. $-\sqrt{1-x}-{\cos }^{-1}\sqrt{x}+\sqrt{1-x}+C$
• D. $-\sqrt{1-x}-2{\cos }^{-1}\sqrt{x}-\sqrt{1-x}+C$

Answer 1

The correct option is B $-2\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}+C$

Let $I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$
Put $x={\cos }^2\theta$
$\Rightarrow dx=-2\cos \theta \sin \theta d\theta$
$\therefore I=-\int \sqrt{\frac{1-\cos \theta }{1+\cos \theta }}2\sin \theta \cos \theta d\theta$

$\Rightarrow I=-\int \frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}\cdot 2\cdot 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\cos \theta d\theta$

$\Rightarrow I=-2\int (1-\cos \theta )\cos \theta d\theta$
$\Rightarrow I=-2\int (\cos \theta -{\cos }^2\theta )d\theta$
$\Rightarrow I=-2\int (\cos \theta -\frac{1+\cos 2\theta }{2})d\theta$
$\Rightarrow I=-2[\sin \theta -\frac{1}{2}(\theta +\frac{\sin 2\theta }{2})]+C$
$\Rightarrow I=-2\sqrt{1-x}+\frac{2}{2}[{\cos }^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}]+C$
$[\text{using}\sin \theta =\sqrt{1-x}]$
$\therefore I=-2\sqrt{1-x}+{\cos }^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}+C$