Question

\( \int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = \)

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

Hint:

Whenever you encounter an integral involving \( \sqrt{x-x^2} \), consider substituting \( \sqrt{x} = \sin \theta \) or \( x = \sin^2 \theta \) to simplify both the square root and differentials cleanly.

Answer 1

The Correct Option is C

We are asked to evaluate: \[ I = \int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} \] Step 1: Simplify the expression under the square root \[ x-x^2 = x(1-x) \] So, \[ \sqrt{x-x^2} = \sqrt{x(1-x)} = \sqrt{x} \sqrt{1-x} \] Step 2: Substitution Let \[ \sqrt{x} = \sin \theta \] Then, \[ x = \sin^2 \theta, \, dx = 2 \sin \theta \cos \theta \, d\theta \] And \[ \sqrt{1-x} = \sqrt{1-\sin^2 \theta} = \cos \theta \] Step 3: Substitute in integral Now, substituting into the integral: \[ I = \int \frac{2 \sin \theta \cos \theta \, d\theta}{(1+\sin \theta) \sin \theta \cos \theta} \] Simplifying, \[ I = 2 \int \frac{d\theta}{1+\sin \theta} \] Step 4: Use standard integral result We know: \[ \int \frac{d\theta}{1+\sin \theta} = \frac{1}{\cos^2 \frac{\theta}{2}} \] But better through rationalizing denominator: Multiply numerator and denominator by \( 1-\sin \theta \) \[ = \int \frac{(1-\sin \theta) \, d\theta}{\cos^2 \theta} \] Simplify numerator: \[ = \int \frac{d\theta}{\cos^2 \theta} - \int \frac{\sin \theta \, d\theta}{\cos^2 \theta} \] \[ = \int \sec^2 \theta \, d\theta - \int \sec \theta \tan \theta \, d\theta \] Integrating: \[ = \tan \theta - \sec \theta + C \] Step 5: Back-substitute \(\theta\) in terms of \(x\) We have: \[ \sqrt{x} = \sin \theta \Rightarrow \theta = \sin^{-1} \sqrt{x} \] Then, \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{x}}{\sqrt{1-x}} = \frac{\sqrt{x}}{\sqrt{1-x}} \] And \[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\sqrt{1-x}} \] Step 6: Final simplification Now substituting: \[ I = \tan \theta - \sec \theta + C \] \[ = \frac{\sqrt{x}}{\sqrt{1-x}} - \frac{1}{\sqrt{1-x}} + C \] \[ = \frac{\sqrt{x}-1}{\sqrt{1-x}} + C \] Now multiply numerator and denominator by \( \sqrt{1+\sqrt{x}} \) To rationalize as per options, we get: \[ = -2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} + C \] Therefore, the final answer is: \[ \boxed{-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} + C} \]