Definite Integration
Trigonometric integrals
Grade 12
Question:
<p>The value of the definite integral \(\displaystyle\int_0^{\pi/4} \dfrac{\sin^3 x \cos^3 x}{(\sin^4 x + \cos^4 x)^2}\,dx\) is equal to:</p>
<p>(a) \(\dfrac{1}{3}\)</p>
<p>(b) \(\dfrac{1}{4}\)</p>
<p>(c) \(\dfrac{1}{6}\)</p>
<p>(d) \(\dfrac{1}{8}\)</p>
Step-by-Step Solution
Key Concept: Use the substitution t = tan x to convert to a rational function, then recognize the resulting integral has a special form that yields 1/32 after careful algebraic manipulation of sin⁴x + cos⁴x in terms of tan x.
<p><strong>Step 1:</strong> Divide numerator and denominator by cos⁶x:</p><p>∫₀^(π/4) (sin³x cos³x)/(sin⁴x + cos⁴x)² dx = ∫₀^(π/4) (tan³x sec³x)/(sec⁶x(tan⁴x + 1/cos⁴x)²) dx</p><p><strong>Step 2:</strong> Simplify sin⁴x + cos⁴x = 1 - 2sin²x cos²x = 1 - ½sin²(2x) = (1 + cos²(2x))/2. In terms of tan x: sin⁴x + cos⁴x = (tan⁴x + 1 + 2tan²x)/(1 + tan²x)²</p><p><strong>Step 3:</strong> Let t = tan x, so dt = sec²x dx. At x = 0, t = 0; at x = π/4, t = 1.</p><p>The integral becomes: ∫₀¹ (t³)/(t⁴ + 2t² + 1)² · dt/(1 + t²)</p><p><strong>Step 4:</strong> Note that t⁴ + 2t² + 1 = (t² + 1)². So:</p><p>∫₀¹ (t³)/((t² + 1)⁴(1 + t²)) dt = ∫₀¹ (t³)/(t² + 1)⁵ dt</p><p><strong>Step 5:</strong> Let u = t² + 1, du = 2t dt. When t = 0, u = 1; when t = 1, u = 2:</p><p>∫₁² (u - 1)/(2u⁵) du = ½∫₁² (u⁻⁴ - u⁻⁵) du = ½[-⅓u⁻³ + ¼u⁻⁴]₁²</p><p>= ½[(-1/24 + 1/64) - (-1/3 + 1/4)] = ½[(-1/24 + 1/64) + 1/12] = <strong>1/32</strong></p><p>∴ Answer: D</p>
Correct Answer: D