Indefinite Integration
Trigonometric Substitution
Grade 12
Question:
<p>Evaluate: \(\int \frac{\sin 2x}{(3 + 4\cos x)^3} dx\)</p>
<p>(A) \(\frac{3 + 8\cos x}{2(3 + 4\cos x)^2} + C\)</p>
<p>(B) \(\frac{3 + 8\cos x}{16(3 + 4\cos x)^2} + C\)</p>
<p>(C) \(\frac{3 - \cos x}{2(3 + 4\cos x)^2} + C\)</p>
<p>(D) \(\frac{3 + 8\cos x}{16(3 + 4\cos x)^2} + C\)</p>
Step-by-Step Solution
Key Concept: Use substitution with $u = 3 + 4\cos x$ and decompose the resulting rational function.
<p><strong>Step 1:</strong> Rewrite $\sin 2x = 2\sin x \cos x$</p><p><strong>Step 2:</strong> Let $u = 3 + 4\cos x$, then $du = -4\sin x dx$</p><p><strong>Step 3:</strong> The integral becomes $-\frac{1}{8} \int \frac{u - 3}{u^3} du$</p><p><strong>Step 4:</strong> Expand and integrate: $-\frac{1}{8}\left[\int u^{-2} du - 3\int u^{-3} du\right]$</p><p><strong>Step 5:</strong> Evaluate to get $\frac{3 + 8\cos x}{16(3 + 4\cos x)^2} + C$</p><p>∴ Answer is B.</p>
Correct Answer: B