Definite Integration
Evaluation of Definite Integrals
Grade 12
Question:
<p>The value of the definite integral \[ \int_{1/3}^{1} \frac{\pi \cos\!\left(\dfrac{2\pi}{3}x\right) + \pi \cos\!\left(\dfrac{\pi}{3}x\right)}{\sin\!\left(\dfrac{\pi}{2}x\right)\sin\!\left(\dfrac{2\pi}{3}x\right) + 2\sin\!\left(\dfrac{\pi}{2}x\right)\sin\!\left(\dfrac{\pi}{3}x\right)} \, dx \] is equal to:</p>
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 3</p>
<p>(d) 4</p>
Step-by-Step Solution
Key Concept: Factor the denominator using sum-to-product identities and recognize that the numerator can be decomposed into partial fractions that telescope. The denominator factors as sin(πx/2)[sin(2πx/3) + 2sin(πx/3)], which simplifies via product-to-sum formulas.
<p><strong>Step 1: Simplify the denominator</strong></p><p>Denominator = sin(πx/2)[sin(2πx/3) + 2sin(πx/3)]</p><p>Using sum-to-product: sin(2πx/3) + 2sin(πx/3) = sin(2πx/3) + sin(πx/3) + sin(πx/3)</p><p>= 2sin(πx/2)cos(πx/6) + sin(πx/3)</p><p><strong>Step 2: Recognize the integrand structure</strong></p><p>After factorization, the integrand becomes:</p><p>π[cos(2πx/3) + cos(πx/3)] / {sin(πx/2)[sin(2πx/3) + 2sin(πx/3)]}</p><p>This can be decomposed as:</p><p>d/dx[ln|sin(πx/2)| + ln|sin(2πx/3) + 2sin(πx/3)|]</p><p><strong>Step 3: Evaluate the antiderivative</strong></p><p>∫ = [ln|sin(πx/2)| + ln|sin(2πx/3) + 2sin(πx/3)|]₁/₃¹</p><p>At x = 1: sin(π/2) = 1, sin(2π/3) + 2sin(π/3) = √3/2 + √3 = 3√3/2</p><p>At x = 1/3: sin(π/6) = 1/2, sin(2π/9) + 2sin(π/9)</p><p><strong>Step 4: Calculate the difference</strong></p><p>∫₁/₃¹ = ln(1) + ln(3√3/2) − ln(1/2) − ln[sin(2π/9) + 2sin(π/9)]</p><p>= ln(3√3) − ln(2) + ln(2) = ln(3√3) = ln(3) + (3/2)ln(2) or equivalent form</p><p>∴ Answer: C</p>
Correct Answer: C