The integral $\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x$ is equal to (where C is the constant of integration)
Answer & explanation
Correct answer: option 2
Let $x=\cos ^2 \theta$. Then, $d x=-2 \sin \theta \cos \theta d \theta$.
∴ $I=\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x$
$\Rightarrow I =\int \frac{1}{(1+\cos \theta) \cos \theta \sin \theta} \times(-2 \sin \theta \cos \theta) d \theta$
$\Rightarrow I=-\int \sec ^2 \frac{\theta}{2} d \theta=-2 \tan \frac{\theta}{2}+C$
$\Rightarrow I=-2 \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}+C=-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C$