Indefinite Integration
Integration of trigonometric functions
Grade 12
Question:
<p>The value of \(\sqrt{2}\int \frac{\sin x\, dx}{\sin\left(x - \frac{\pi}{4}\right)}\) is</p>
<p>\(x + \log\left|\cos\left(x - \frac{\pi}{4}\right)\right| + C\)</p>
<p>\(x - \log\left|\sin\left(x - \frac{\pi}{4}\right)\right| + C\)</p>
<p>\(x + \log\left|\sin\left(x - \frac{\pi}{4}\right)\right| + C\)</p>
<p>\(x - \log\left|\cos\left(x - \frac{\pi}{4}\right)\right| + C\)</p>
Step-by-Step Solution
Key Concept: Expand sin(x - π/4) using sine difference formula and rewrite the integrand to separate it into standard forms. The denominator becomes (sin x - cos x)/√2, allowing the integral to decompose into logarithmic and arctangent terms.
<p><strong>Step 1:</strong> Expand the denominator using sine difference formula:</p><p>sin(x - π/4) = sin x cos(π/4) - cos x sin(π/4) = (sin x - cos x)/√2</p><p><strong>Step 2:</strong> Rewrite the integral:</p><p>√2 ∫ sin x dx / [(sin x - cos x)/√2] = √2 · √2 ∫ sin x dx/(sin x - cos x) = 2∫ sin x dx/(sin x - cos x)</p><p><strong>Step 3:</strong> Express numerator in useful form. Let sin x = A(sin x - cos x) + B(-cos x + sin x)·(-1)</p><p>Rewrite: sin x = A(sin x - cos x) + B(cos x - sin x), so sin x = (A - B)sin x + (B - A)cos x</p><p>Thus: A - B = 1 and B - A = 0, giving A = 1/2, B = -1/2</p><p><strong>Step 4:</strong> Split the integral:</p><p>2∫ [(1/2)(sin x - cos x) - (1/2)(cos x - sin x)]/(sin x - cos x) dx</p><p>= 2∫ [1/2 - (1/2)(cos x - sin x)/(sin x - cos x)] dx</p><p>= 2∫ [1/2 + (1/2)] dx - ∫ (cos x - sin x)/(sin x - cos x) dx</p><p>= ∫ dx + ∫ d(sin x - cos x)/(sin x - cos x)</p><p><strong>Step 5:</strong> Integrate:</p><p>= x + ln|sin x - cos x| + C</p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C