Indefinite Integration
Integration by parts
Grade 12
Question:
<p>Given \(I = \int \cos(\log_e x)\, dx\). Then \(I\) equals:</p>
<p>\(x[\sin(\log_e x) - \cos(\log_e x)] + C\)</p>
<p>\(\dfrac{x}{2}[\cos(\log_e x) - \sin(\log_e x)] + C\)</p>
<p>\(\dfrac{x}{2}[\sin(\log_e x) + \cos(\log_e x)] + C\)</p>
<p>\(x[\sin(\log_e x) + \cos(\log_e x)] + C\)</p>
Step-by-Step Solution
Key Concept: Use integration by parts twice on cos(log x), then solve the resulting recursive equation for I. The key is recognizing that the integral reappears on the right side, allowing you to collect terms and isolate I.
<p><strong>Step 1:</strong> Apply integration by parts with u = cos(log x), dv = dx</p><p>Then du = -sin(log x)·(1/x) dx, v = x</p><p>I = x·cos(log x) + ∫ sin(log x) dx</p><p><strong>Step 2:</strong> Apply integration by parts again to ∫ sin(log x) dx with u = sin(log x), dv = dx</p><p>Then du = cos(log x)·(1/x) dx, v = x</p><p>∫ sin(log x) dx = x·sin(log x) - ∫ cos(log x) dx = x·sin(log x) - I</p><p><strong>Step 3:</strong> Substitute back into the equation from Step 1</p><p>I = x·cos(log x) + x·sin(log x) - I</p><p><strong>Step 4:</strong> Solve for I by collecting terms</p><p>2I = x·cos(log x) + x·sin(log x)</p><p>I = (x/2)[cos(log x) + sin(log x)] + C</p><p>∴ Answer: C</p>
Correct Answer: C