Definite Integration
Evaluation of Definite Integrals by Substitution
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 the integral into a rational function, then recognize that the denominator simplifies to (1 + t^4)^2 after factoring out appropriate powers. The numerator becomes t^3 dt, making this a standard arctangent-related integral.
<p><strong>Step 1:</strong> Divide numerator and denominator by cos^6 x:</p><p>$$I = \int_0^{\pi/4} \frac{\tan^3 x \sec^6 x}{(\tan^4 x + 1)^2 \sec^8 x}\, dx = \int_0^{\pi/4} \frac{\tan^3 x}{(\tan^4 x + 1)^2}\sec^2 x\, dx$$</p><p><strong>Step 2:</strong> Let t = tan x, so dt = sec² x dx. When x = 0, t = 0; when x = π/4, t = 1:</p><p>$$I = \int_0^1 \frac{t^3}{(t^4 + 1)^2}\, dt$$</p><p><strong>Step 3:</strong> Use substitution u = t^4 + 1, so du = 4t³ dt:</p><p>$$I = \frac{1}{4}\int_1^2 \frac{du}{u^2} = \frac{1}{4}\left[-\frac{1}{u}\right]_1^2 = \frac{1}{4}\left(-\frac{1}{2} + 1\right) = \frac{1}{4} \cdot \frac{1}{2}$$</p><p><strong>Step 4:</strong> $$I = \frac{1}{8}$$</p><p>∴ Answer: D</p>
Correct Answer: D