Indefinite Integration
Integration of trigonometric functions
Grade 12
Question:
<p>\(\int \frac{\sin^8 x - \cos^8 x}{1-2\sin^2 x\cos^2 x}dx\) is equal to</p>
<p>\(\frac{1}{2}\sin 2x + C\)</p>
<p>\(-\frac{1}{2}\sin 2x + C\)</p>
<p>\(-\frac{1}{2}\sin x + C\)</p>
<p>\(-\sin^2 x + C\)</p>
Step-by-Step Solution
Key Concept: Factor the numerator as a difference of fourth powers: sin⁸x - cos⁸x = (sin⁴x - cos⁴x)(sin⁴x + cos⁴x), then use the identity 1 - 2sin²x cos²x = sin⁴x + cos⁴x to simplify the fraction to sin⁴x - cos⁴x.
<p><strong>Step 1:</strong> Factor the numerator using difference of fourth powers: sin⁸x - cos⁸x = (sin⁴x - cos⁴x)(sin⁴x + cos⁴x)</p><p><strong>Step 2:</strong> Recognize that the denominator 1 - 2sin²x cos²x can be rewritten. Note that sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x cos²x = 1 - 2sin²x cos²x</p><p><strong>Step 3:</strong> Cancel the common factor: ∫[(sin⁴x - cos⁴x)(sin⁴x + cos⁴x)]/(sin⁴x + cos⁴x) dx = ∫(sin⁴x - cos⁴x)dx</p><p><strong>Step 4:</strong> Factor further: sin⁴x - cos⁴x = (sin²x - cos²x)(sin²x + cos²x) = (sin²x - cos²x)(1) = -cos(2x)</p><p><strong>Step 5:</strong> Integrate: ∫(-cos 2x)dx = -(sin 2x)/2 + C = <strong>-sin(2x)/2 + C</strong></p><p>∴ Answer: B</p>
Correct Answer: B