Indefinite Integration
Integration by substitution/reduction
Grade 12
Question:
<p>We have
\[I = \int \frac{\sin x}{\sin(x - \alpha)} dx\]
If \(I = Ax + B\ln|\sin(x-\alpha)| + C\), then find the values of \(A\) and \(B\).</p>
<p>\(A = \cos\alpha,\ B = \sin\alpha\)</p>
<p>\(A = \sin\alpha,\ B = \cos\alpha\)</p>
<p>\(A = -\cos\alpha,\ B = \sin\alpha\)</p>
<p>\(A = \cos\alpha,\ B = -\sin\alpha\)</p>
Step-by-Step Solution
Key Concept: Decompose the numerator sin x as sin[(x-α)+α] = sin(x-α)cos α + cos(x-α)sin α to split the integral into recognizable parts with one term matching the denominator's derivative.
<p><strong>Step 1:</strong> Decompose sin x using the angle addition formula:</p><p>sin x = sin[(x-α)+α] = sin(x-α)cos α + cos(x-α)sin α</p><p><strong>Step 2:</strong> Split the integral:</p><p>I = ∫[sin(x-α)cos α + cos(x-α)sin α]/sin(x-α) dx</p><p>I = ∫cos α dx + ∫[cos(x-α)sin α]/sin(x-α) dx</p><p><strong>Step 3:</strong> Evaluate the first integral:</p><p>∫cos α dx = (cos α)x = (cos α)x + C₁</p><p><strong>Step 4:</strong> For the second integral, note that d/dx[sin(x-α)] = cos(x-α):</p><p>∫[cos(x-α)sin α]/sin(x-α) dx = sin α · ∫cos(x-α)/sin(x-α) dx</p><p>= sin α · ln|sin(x-α)| + C₂</p><p><strong>Step 5:</strong> Combine both parts:</p><p>I = (cos α)x + (sin α)ln|sin(x-α)| + C</p><p><strong>Step 6:</strong> Compare with I = Ax + B ln|sin(x-α)| + C:</p><p>∴ <strong>A = cos α</strong> and <strong>B = sin α</strong></p>
Correct Answer: A