Question

\(\int _{0}^{4}\sqrt{\frac{4-x}{4+x}}dx=\)

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

Hint:

Substitute x = 4 cos theta so the root becomes tan(theta/2).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The form \(\sqrt{\dfrac{a-x}{a+x}}\) suggests \(x=a\cos \theta\).

Step 2: Substitute
Let \(x=4\cos \theta\). Then \(\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}=\tan \dfrac{\theta }{2}\) and \(dx=-4\sin \theta \,d\theta\).
Limits: \(x=0\) gives \(\theta =\dfrac{\pi }{2}\); \(x=4\) gives \(\theta =0\).

Step 3: Simplify
\(I=\int _{0}^{\pi /2}\tan \dfrac{\theta }{2}\cdot 4\sin \theta \,d\theta =\int _{0}^{\pi /2}8\sin ^{2}\dfrac{\theta }{2}\,d\theta =4\int _{0}^{\pi /2}(1-\cos \theta )d\theta\).

Step 4: Evaluate
\(I=4\left(\dfrac{\pi }{2}-1\right)=2\pi -4\).

Final Answer:
The integral equals \(2\pi -4\). \[ \boxed{\text{(A) }2\pi -4} \]