<p>The value of the definite integral \[I = \int_0^{\pi/2} \frac{\cos^4 x + \sin x\cos^3 x + \sin^2 x\cos^2 x + \sin^3 x\cos x}{\sin^4 x + \cos^4 x + 2\sin x\cos^3 x + 2\sin^2 x\cos^2 x + 2\sin^3 x\cos x}\, dx\] is equal to \(\frac{\pi}{k}\). Find \(k\).</p>
Step-by-Step Solution
Key Concept: Factor the numerator as cos x(cos³x + sin cos²x + sin x cos x + sin³x) and denominator as (sin²x + cos²x)² = 1, then recognize the numerator factors to (sin x + cos x)·cos x·(sin x + cos x) = cos x(sin x + cos x)². Use substitution u = sin x + cos x to evaluate.
<p><strong>Step 1:</strong> Rewrite the denominator. Notice that sin⁴x + cos⁴x + 2sin x cos³x + 2sin²x cos²x + 2sin³x cos x = (sin²x + cos²x)² + 2sin x cos x(sin²x + cos²x) = 1 + 2sin x cos x = (sin x + cos x)².</p><p><strong>Step 2:</strong> Factor the numerator: cos⁴x + sin x cos³x + sin²x cos²x + sin³x cos x = cos x(cos³x + sin cos²x + sin x cos x + sin³x) = cos x[(cos³x + sin³x) + sin x cos x(cos x + sin x)] = cos x(sin x + cos x)(cos²x - sin x cos x + sin²x + sin x cos x) = cos x(sin x + cos x).</p><p><strong>Step 3:</strong> The integral becomes:</p><p>I = ∫₀^(π/2) [cos x(sin x + cos x)]/[(sin x + cos x)²] dx = ∫₀^(π/2) cos x/(sin x + cos x) dx</p><p><strong>Step 4:</strong> Substitute u = sin x + cos x, so du = (cos x - sin x) dx. Note that cos x = (1/2)[du/dx + (u² - 1)/(2u)]. Alternatively, use the fact that d/dx[ln(sin x + cos x)] = (cos x - sin x)/(sin x + cos x).</p><p><strong>Step 5:</strong> Let u = sin x + cos x. Then du = (cos x - sin x) dx. We need ∫ cos x/(sin x + cos x) dx. Write cos x = (1/2)[(cos x - sin x) + (sin x + cos x)] = (1/2)(du/dx + u) dx. This gives: I = (1/2)ln(sin x + cos x)|₀^(π/2) + (1/2)∫₀^(π/2) 1 dx = (1/2)ln(1) - (1/2)ln(1) + π/4 = π/4.</p><p><strong>Step 6:</strong> Since I = π/4 = π/k, we have k = 4.</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4