Question

The integral $\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$ is equal to (where $C$ is a constant of integration)

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

Answer 1

Answer: (B)

$I=\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$
put $x={\cos }^2\theta$
$dx=-2\cos \theta \sin \theta d\theta$
$I=\int \frac{-2\sin \theta \cos \theta d\theta }{(1+\cos \theta )\cos \theta \sin \theta }=-2\int \frac{d\theta }{2{\cos }^2\theta /2}$
$=-\int {\sec }^2\left(\frac{\theta }{2}\right)d\theta \therefore \cos \theta =\sqrt{x}$
$=-2\tan \theta /2+C\frac{1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}=\sqrt{x}$
$=-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+c$
$\Rightarrow {\tan }^2\left(\frac{\theta }{2}\right)=\frac{1-\sqrt{x}}{1+\sqrt{x}}$