Definite Integration
Evaluation of definite integrals
Grade 12
Question:
<p>The value of
\[\frac{\int_0^{\pi/2}(5\cos^2 x+3\sin^2 x)\,dx}{\int_0^{\pi/2}\sin\theta\cos\theta\sqrt{25\sin^2\theta+9\cos^2\theta}\,d\theta}\]
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 by expressing 5cos²x + 3sin²x = 2cos²x + 3(sin²x + cos²x) = 2cos²x + 3, then use the substitution u = 5sin²θ + 9cos²θ in the denominator to transform it into a recognizable form.
<p><strong>Step 1: Simplify the numerator</strong></p><p>5cos²x + 3sin²x = 5cos²x + 3sin²x = 2cos²x + 3(sin²x + cos²x) = 2cos²x + 3</p><p>∫₀^(π/2) (2cos²x + 3)dx = 2·(π/4) + 3·(π/2) = π/2 + 3π/2 = 2π</p><p><strong>Step 2: Evaluate the denominator using substitution</strong></p><p>Let u = 25sin²θ + 9cos²θ</p><p>Then du = (50sinθcosθ - 18cosθsinθ)dθ = 32sinθcosθ dθ</p><p>So sinθcosθ dθ = du/32</p><p>When θ = 0: u = 9; when θ = π/2: u = 25</p><p>∫₀^(π/2) sinθcosθ√(25sin²θ + 9cos²θ) dθ = ∫₉²⁵ √u · (du/32) = (1/32)·[⅔u^(3/2)]₉²⁵</p><p>= (1/48)[125 - 27] = 98/48 = 49/24</p><p><strong>Step 3: Find the ratio</strong></p><p>2π ÷ (49/24) = 2π · (24/49) = 48π/49</p><p>∴ Answer: B</p>
Correct Answer: B