Question

The value of $\int_{0}^{4} \sqrt{\frac{4-x}{4+x}} dx$ is:

• A. $4(\pi - 2)$
• B. $2(\pi - 2)$
• C. $4(\pi + 2)$
• D. $2(\pi + 2)$

Hint:

For integrals involving $\sqrt{\frac{a-x}{a+x}}$, rationalizing the numerator or using $x = a \cos \theta$ are effective strategies.

Answer 1

The Correct Option is B

Step 1: Concept
To evaluate the integral, we use trigonometric substitution to simplify the square root expression. A common substitution for $\sqrt{\frac{a-x}{a+x}}$ is $x = a \cos \theta$.

Step 2: Meaning
Let $x = 4 \cos \theta$. Then $dx = -4 \sin \theta d\theta$. When $x = 0$, $\theta = \pi/2$; when $x = 4$, $\theta = 0$.

Step 3: Analysis
Substituting these into the integral: $\int_{\pi/2}^{0} \sqrt{\frac{4(1-\cos \theta)}{4(1+\cos \theta)}} (-4 \sin \theta) d\theta = \int_{0}^{\pi/2} \sqrt{\frac{2 \sin^2(\theta/2)}{2 \cos^2(\theta/2)}} (4 \sin \theta) d\theta$. This simplifies to $\int_{0}^{\pi/2} \tan(\theta/2) \cdot 8 \sin(\theta/2) \cos(\theta/2) d\theta = \int_{0}^{\pi/2} 8 \sin^2(\theta/2) d\theta = \int_{0}^{\pi/2} 4(1 - \cos \theta) d\theta$.

Step 4: Conclusion
Integrating: $4[\theta - \sin \theta]_{0}^{\pi/2} = 4(\pi/2 - 1) = 2\pi - 4 = 2(\pi - 2)$. Final Answer: (B)