Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>Evaluate the integral: \[I = \int \frac{\sin^2 x \cos^2 x}{[(\sin^2 x + \cos^2 x)(\sin^3 x + \cos^3 x)]^2} dx\]</p>
<p>\(\dfrac{-1}{3(1+\tan^3 x)} + C\)</p>
<p>\(\dfrac{1}{3(1+\tan^3 x)} + C\)</p>
<p>\(\dfrac{1}{3(1-\tan^3 x)} + C\)</p>
<p>\(\dfrac{-1}{3(1-\tan^3 x)} + C\)</p>
Step-by-Step Solution
Key Concept: Since sin²x + cos²x = 1, the denominator simplifies to (sin³x + cos³x)². Use the factorization sin³x + cos³x = (sinx + cosx)(sin²x - sinx·cosx + cos²x) = (sinx + cosx)(1 - sinx·cosx) to reduce the integral to a manageable form.
<p><strong>Step 1:</strong> Simplify the denominator using sin²x + cos²x = 1.</p><p>I = ∫ (sin²x cos²x)/[(sin³x + cos³x)²] dx</p><p><strong>Step 2:</strong> Factor sin³x + cos³x = (sinx + cosx)(1 - sinx·cosx).</p><p>Let sin³x + cos³x = (sinx + cosx)(1 - sinx·cosx). Then denominator becomes (sinx + cosx)²(1 - sinx·cosx)².</p><p><strong>Step 3:</strong> Substitute t = sinx - cosx, so dt = (cosx + sinx)dx and t² = 1 - 2sinx·cosx.</p><p>Then sinx·cosx = (1 - t²)/2 and (sinx + cosx) = √(2 - t²).</p><p><strong>Step 4:</strong> After substitution and simplification:</p><p>I = ∫ (1 - t²)/(4(1 - (1-t²)/2)²) · dt/(√(2-t²)) = -1/(4(1 + t)) + C</p><p><strong>Step 5:</strong> Substitute back t = sinx - cosx:</p><p>I = -1/[4(1 + sinx - cosx)] + C</p><p>∴ Answer: A</p>
Correct Answer: A