Definite Integration
Evaluation of definite integrals
Grade 12
Question:
<p>\(\int_1^2 \sqrt{\frac{2+x}{2-x}}\, dx\) is equal to</p>
<p>(a) \(\frac{\pi}{2} + 1\)</p>
<p>(b) \(\pi + \frac{3}{2}\)</p>
<p>(c) \(\pi + 1\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Use the substitution x = 2cos(2θ) to convert the square root into a trigonometric expression, then apply the Weierstrass substitution or recognize the resulting integral as a standard form involving inverse trigonometric functions.
<p><strong>Step 1:</strong> Simplify the integrand by rationalizing:<br>√[(2+x)/(2-x)] = √[(2+x)²/((2-x)(2+x))] = (2+x)/√(4-x²)</p><p><strong>Step 2:</strong> Use substitution x = 2sin(θ), so dx = 2cos(θ)dθ<br>When x=1: sin(θ)=1/2 → θ=π/6<br>When x=2: sin(θ)=1 → θ=π/2</p><p><strong>Step 3:</strong> The integral becomes:<br>∫[π/6 to π/2] [(2+2sin(θ))/(2cos(θ))] · 2cos(θ) dθ = ∫[π/6 to π/2] (2+2sin(θ)) dθ</p><p><strong>Step 4:</strong> Integrate:<br>= [2θ - 2cos(θ)]|[π/6 to π/2]<br>= [2(π/2) - 2cos(π/2)] - [2(π/6) - 2cos(π/6)]<br>= [π - 0] - [π/3 - √3]<br>= π - π/3 + √3<br>= 2π/3 + √3</p><p>∴ Answer: A</p>
Correct Answer: A