Definite Integration
Evaluation of Definite Integrals
Grade 12
Question:
<p>The integral \(\displaystyle\int_{\pi/2}^{\pi/4} \dfrac{8\cos 2x}{(\tan x + \cot x)^3}\,dx\) equals</p>
<p>\(\dfrac{15}{128}\)</p>
<p>\(\dfrac{13}{32}\)</p>
<p>\(\dfrac{13}{256}\)</p>
<p>\(\dfrac{15}{64}\)</p>
Step-by-Step Solution
Key Concept: Simplify the denominator using the identity (tan x + cot x) = sin(2x)/sin(x)cos(x), then recognize that after substitution u = sin(2x), the integral becomes a standard power function integral.
<p><strong>Step 1:</strong> Simplify the denominator. We have:</p><p>tan x + cot x = (sin x/cos x) + (cos x/sin x) = (sin²x + cos²x)/(sin x cos x) = 1/(sin x cos x) = 2/sin(2x)</p><p><strong>Step 2:</strong> Therefore (tan x + cot x)³ = 8/sin³(2x)</p><p><strong>Step 3:</strong> Substitute into the integral:</p><p>∫[π/2 to π/4] (8cos(2x))/(8/sin³(2x)) dx = ∫[π/2 to π/4] cos(2x)·sin³(2x) dx</p><p><strong>Step 4:</strong> Let u = sin(2x), so du = 2cos(2x)dx</p><p>When x = π/2: u = sin(π) = 0</p><p>When x = π/4: u = sin(π/2) = 1</p><p><strong>Step 5:</strong> The integral becomes:</p><p>∫[0 to 1] u³·(du/2) = (1/2)∫[0 to 1] u³ du = (1/2)·[u⁴/4]₀¹ = (1/2)·(1/4) = 1/8</p><p>∴ Answer: <strong>C</strong> (or 1/8)</p>
Correct Answer: C