<p>If \(\displaystyle\int_0^{\pi/3} \frac{\tan\theta}{\sqrt{2k\sec\theta}}\, d\theta = 1 - \dfrac{1}{\sqrt{2}}\), \((k > 0)\), then the value of \(k\) is:</p>
Step-by-Step Solution
Key Concept: Rewrite the integrand using sec θ = 1/cos θ and tan θ = sin θ/cos θ, then substitute u = cos θ to convert to a power function integral that yields a solvable algebraic equation.
<p><strong>Step 1:</strong> Simplify the integrand. Rewrite:</p><p>$$\frac{\tan\theta}{\sqrt{2k\sec\theta}} = \frac{\sin\theta/\cos\theta}{\sqrt{2k/\cos\theta}} = \frac{\sin\theta}{\sqrt{2k}\sqrt{\cos\theta}}$$</p><p><strong>Step 2:</strong> Use substitution u = cos θ, so du = −sin θ dθ. When θ = 0, u = 1; when θ = π/3, u = 1/2.</p><p>$$\int_0^{\pi/3} \frac{\sin\theta}{\sqrt{2k}\sqrt{\cos\theta}}\, d\theta = \int_1^{1/2} \frac{-1}{\sqrt{2k}\sqrt{u}}\, du = \frac{1}{\sqrt{2k}}\int_{1/2}^1 u^{-1/2}\, du$$</p><p><strong>Step 3:</strong> Evaluate the integral:</p><p>$$\frac{1}{\sqrt{2k}}\left[2\sqrt{u}\right]_{1/2}^1 = \frac{2}{\sqrt{2k}}\left(1 - \frac{1}{\sqrt{2}}\right) = 1 - \frac{1}{\sqrt{2}}$$</p><p><strong>Step 4:</strong> Solve for k:</p><p>$$\frac{2}{\sqrt{2k}} = 1 \implies \sqrt{2k} = 2 \implies 2k = 4 \implies k = 2$$</p><p>∴ Answer: D</p>
Correct Answer: D