Question

Evaluate \[ \int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}. \]

• 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:

For integrals containing \(\sqrt{x-x^2}\), the substitution \(\sqrt{x}=\sin\theta\) is usually effective because \[ x-x^2=\sin^2\theta\cos^2\theta. \] This converts the radical into a simple trigonometric product.

Answer 1

The Correct Option is C

Concept: Use the substitution \[ \sqrt{x}=\sin\theta. \] This transforms the radical \(\sqrt{x-x^2}\) into a simple trigonometric expression.

Step 1: Substitute \(\sqrt{x}=\sin\theta\). Then \[ x=\sin^2\theta, \] and \[ dx=2\sin\theta\cos\theta\,d\theta. \] Also, \[ \sqrt{x-x^2} = \sqrt{\sin^2\theta-\sin^4\theta} = \sin\theta\cos\theta. \] Therefore, \[ I = \int \frac{2\sin\theta\cos\theta\,d\theta} {(1+\sin\theta)(\sin\theta\cos\theta)}. \] \[ = 2\int\frac{d\theta}{1+\sin\theta}. \]

Step 2: Simplify the integrand. \[ I = 2\int \frac{1-\sin\theta}{1-\sin^2\theta} \,d\theta. \] \[ = 2\int \frac{1-\sin\theta}{\cos^2\theta} \,d\theta. \] \[ = 2\int \left(\sec^2\theta-\sec\theta\tan\theta\right) d\theta. \] \[ = 2\left(\tan\theta-\sec\theta\right)+C. \]

Step 3: Express the result in terms of \(x\). Since \[ \sin\theta=\sqrt{x}, \] \[ \tan\theta = \frac{\sqrt{x}}{\sqrt{1-x}}, \] and \[ \sec\theta = \frac{1}{\sqrt{1-x}}. \] Hence, \[ I = 2\left( \frac{\sqrt{x}-1}{\sqrt{1-x}} \right)+C. \] \[ = -2 \left( \frac{1-\sqrt{x}} {\sqrt{(1-\sqrt{x})(1+\sqrt{x})}} \right)+C. \] \[ = -2 \sqrt{ \frac{1-\sqrt{x}} {1+\sqrt{x}} } +C. \] Therefore, \[ \boxed{ \int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}} = -2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C } \] \[ \boxed{\text{Answer = (C)}} \]