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

The correct option is A $-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$

$I=\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$
Put $x={\cos }^2\theta dx=-2\cos \theta \sin \theta d\theta I=\int \frac{-2\cos \theta \sin \theta d\theta }{(1+\cos \theta )\cos \theta \sin \theta }=-2\int \frac{d\theta }{2{\cos }^2\frac{\theta }{2}}=-\int {\sec }^2\left(\frac{\theta }{2}\right)d\theta =-2\tan \frac{\theta }{2}+C\cos \theta =\sqrt{x},\text{and}\tan \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}I=-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$