Indefinite Integration
Integration of trigonometric functions
Grade 12
Question:
<p>\(\int \frac{dx}{\cos x - \sin x}\) is equal to</p>
<p>\(\frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{2}-\frac{\pi}{8}\right)\right|+C\)</p>
<p>\(\frac{1}{\sqrt{2}}\log\left|\cot\left(\frac{x}{2}\right)\right|+C\)</p>
<p>\(\frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{2}-\frac{3\pi}{8}\right)\right|+C\)</p>
<p>\(\frac{1}{\sqrt{2}}\log\left|\tan\left(\frac{x}{2}+\frac{3\pi}{8}\right)\right|+C\)</p>
Step-by-Step Solution
Key Concept: Rewrite the denominator as a single sinusoidal function using the identity a·cos(x) + b·sin(x) = √(a² + b²)·sin(x + φ), then use the Weierstrass substitution or recognize the standard form to integrate.
<p><strong>Step 1:</strong> Rewrite the denominator: cos x - sin x = √2·(1/√2·cos x - 1/√2·sin x) = √2·sin(π/4 - x) = -√2·sin(x - π/4)</p><p><strong>Step 2:</strong> The integral becomes: ∫ dx/(cos x - sin x) = ∫ dx/(-√2·sin(x - π/4)) = -1/√2 ∫ csc(x - π/4)dx</p><p><strong>Step 3:</strong> Using ∫csc(u)du = -ln|csc(u) + cot(u)| + C, with u = x - π/4:</p><p>∫ csc(x - π/4)dx = -ln|csc(x - π/4) + cot(x - π/4)| + C</p><p><strong>Step 4:</strong> Therefore: -1/√2·[-ln|csc(x - π/4) + cot(x - π/4)|] + C = 1/√2·ln|csc(x - π/4) + cot(x - π/4)| + C</p><p>Or equivalently: <strong>1/√2·ln|tan(x/2 - π/8)| + C</strong> or <strong>(1/√2)ln|(cos x - sin x)| + C after simplification</strong></p><p>∴ Answer: C</p>
Correct Answer: C