Indefinite Integration
Trigonometric Integration
Grade 12
Question:
<p><span class="math">\(\int (\sin 2x - \cos 2x) \, dx = \frac{1}{2}\sin(2x - a) + b\)</span>, then</p>
<p>(a) <span class="math">\(a = \frac{5\pi}{4}, b \in \mathbb{R}\)</span></p>
<p>(b) <span class="math">\(a = -\frac{5\pi}{4}, b \in \mathbb{R}\)</span></p>
<p>(c) <span class="math">\(a = \frac{\pi}{4}, b \in \mathbb{R}\)</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Convert the sum of sine and cosine into a single trigonometric function using the formula a·sin(θ) + b·cos(θ).
<p><strong>Solution:</strong> Rewrite <span class="math">$\sin 2x - \cos 2x = \sqrt{2}\sin(2x - \frac{\pi}{4})$</span>. Thus <span class="math">$\int (\sin 2x - \cos 2x) \, dx = -\frac{\sqrt{2}}{2}\cos(2x - \frac{\pi}{4}) + C = \frac{\sqrt{2}}{2}\sin(2x - \frac{\pi}{4} - \frac{\pi}{2}) + C = \frac{\sqrt{2}}{2}\sin(2x - \frac{5\pi}{4}) + C$</span>. Comparing with <span class="math">$\frac{1}{2}\sin(2x - a) + b$</span> gives <span class="math">$a = \frac{5\pi}{4}$</span>.</p>
Correct Answer: A