Question

Evaluate the integral: \[ \int_{0}^{1} \sqrt{\frac{2 + x}{2 - x}} \, dx \]

• A. \( \pi + 2 \)
• B. \( \frac{1}{2} (\pi + 2) \)
• C. \( \frac{\pi}{2} + 2 + \sqrt{3} \)
• D. \( \frac{\pi}{3} 2 - \sqrt{3} \)

Hint:

For integrals involving square roots of rational expressions, trigonometric substitutions can be useful to simplify the given integral.

Answer 1

The Correct Option is D

Step 1: Substituting a Trigonometric Identity
Let: \[ x = 2 \sin \theta, \quad dx = 2 \cos \theta \, d\theta. \] Rewriting the integral in terms of \( \theta \): \[ I = \int \sqrt{\frac{2 + 2\sin\theta}{2 - 2\sin\theta}} \cdot 2 \cos\theta \, d\theta. \] Step 2: Simplifying the Expression
\[ I = \int \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} \cdot 2 \cos\theta \, d\theta. \] Using the identity: \[ \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = \tan\left(\frac{\pi}{4} + \frac{\theta}{2} \right), \] \[ I = \int 2 \tan\left(\frac{\pi}{4} + \frac{\theta}{2} \right) \cos\theta \, d\theta. \] Splitting the integral: \[ I = 2 \int \tan\left(\frac{\pi}{4} + \frac{\theta}{2} \right) \cos\theta \, d\theta. \] Step 3: Evaluating the Integral
After integration and simplification: \[ I = \frac{\pi}{3} + 2 - \sqrt{3}. \] Step 4: Conclusion
Thus, the final answer is: \[ \boxed{\frac{\pi}{3} + 2 - \sqrt{3}}. \]