Indefinite Integration
Trigonometric Functions
Grade 12
Question:
<p>Evaluate: \(I = \int \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} dx\)</p>
<p>(a) \(\sin 2x + C\)</p>
<p>(b) \(\frac{\sin^2 x}{2} + C\)</p>
<p>(c) \(-\frac{\sin^2 x}{2} + C\)</p>
<p>(d) \(-2\sin 2x + C\)</p>
Step-by-Step Solution
Key Concept: Factor the numerator using difference of squares and recognize standard trigonometric identities to simplify the integrand.
<p><strong>Step 1:</strong> Factor the numerator: $\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$. Further factor: $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = (\sin^2 x - \cos^2 x)$. </p><p><strong>Step 2:</strong> Express in terms of $\sin^2 x + \cos^2 x = 1$ and double angle identities. The denominator $1 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$. </p><p><strong>Step 3:</strong> Simplify and integrate using substitution or recognizing standard forms. The result is $-\frac{\sin^2 x}{2} + C$.</p>
Correct Answer: C