Definite Integration
Trigonometric substitution in definite integrals
Grade 12
Question:
<p>Given \( \int_0^{\pi/3} \dfrac{\tan\theta}{\sqrt{2k\sec\theta}} \, d\theta = 1 - \dfrac{1}{\sqrt{2}} \), find the value of \( k \).</p>
Step-by-Step Solution
Key Concept: Rewrite the integrand using sec θ = 1/cos θ and tan θ = sin θ/cos θ, then use substitution u = cos θ to transform this into a standard form that evaluates to a power function.
<p><strong>Step 1:</strong> Rewrite the integrand:</p><p>∫₀^(π/3) tan θ/√(2k sec θ) dθ = ∫₀^(π/3) (sin θ/cos θ)/√(2k/cos θ) dθ</p><p>= ∫₀^(π/3) (sin θ/cos θ) · √(cos θ/2k) dθ = (1/√(2k)) ∫₀^(π/3) sin θ/√(cos θ) dθ</p><p><strong>Step 2:</strong> Use substitution u = cos θ, so du = -sin θ dθ</p><p>When θ = 0: u = 1; when θ = π/3: u = 1/2</p><p>= (1/√(2k)) ∫₁^(1/2) (-1/√u) du = (1/√(2k)) ∫_(1/2)^1 u^(-1/2) du</p><p><strong>Step 3:</strong> Evaluate the integral:</p><p>= (1/√(2k)) [2√u]_(1/2)^1 = (1/√(2k)) [2(1) - 2(1/√2)]</p><p>= (1/√(2k)) [2 - √2] = (2 - √2)/√(2k)</p><p><strong>Step 4:</strong> Set equal to given value:</p><p>(2 - √2)/√(2k) = 1 - 1/√2 = (√2 - 1)/√2</p><p>Cross-multiply: √2(2 - √2) = (√2 - 1)√(2k)</p><p>2√2 - 2 = (√2 - 1)√(2k)</p><p>√(2k) = (2√2 - 2)/(√2 - 1) = 2(√2 - 1)/(√2 - 1) = 2</p><p>2k = 4</p><p>∴ <strong>k = 2</strong></p>
Correct Answer: 2