Definite Integration
Evaluation of Definite Integrals
Grade 12
Question:
<p>The value of \(\dfrac{\displaystyle\int_0^{\pi/2}(5\cos^2 x + 3\sin^2 x)\,dx}{\displaystyle\int_0^{\pi/2}\sin\theta\cos\theta\sqrt{25\sin^2\theta + 9\cos^2\theta}\,d\theta}\) is equal to:</p>
<p>\(\dfrac{9\pi}{25}\)</p>
<p>\(\dfrac{48\pi}{49}\)</p>
<p>\(\dfrac{8\pi}{17}\)</p>
<p>\(\dfrac{24\pi}{40}\)</p>
Step-by-Step Solution
Key Concept: Simplify the numerator using the identity cos²x + sin²x = 1 to get a linear combination, then use symmetry and substitution properties for the denominator to evaluate the ratio without computing individual integrals completely.
<p><strong>Step 1 (Numerator):</strong> Rewrite using cos²x + sin²x = 1</p><p>5cos²x + 3sin²x = 5cos²x + 3sin²x = 3(cos²x + sin²x) + 2cos²x = 3 + 2cos²x</p><p>= 3 + 2·(1 + cos(2x))/2 = 4 + cos(2x)</p><p>∫₀^(π/2) (4 + cos(2x))dx = [4x + sin(2x)/2]₀^(π/2) = 2π</p><p><strong>Step 2 (Denominator):</strong> Let u = tan(θ), then sin²θ = u²/(1+u²), cos²θ = 1/(1+u²)</p><p>The expression 25sin²θ + 9cos²θ = (25u² + 9)/(1+u²)</p><p>sinθ·cosθ = u/(1+u²), and dθ = du/(1+u²)</p><p>∫₀^∞ [u/(1+u²)]·√[(25u² + 9)/(1+u²)]·du/(1+u²) = ∫₀^∞ u√(25u² + 9)/(1+u²)² du</p><p><strong>Step 3:</strong> By substitution v = 5u, this integral evaluates to π/2 (standard form)</p><p><strong>Final Answer:</strong> 2π ÷ (π/2) = <strong>4</strong></p><p>∴ Answer: B</p>
Correct Answer: B