Indefinite Integration
Integration by Parts
Grade 12
Question:
<p>\(\int \sin^2(\ln x) \, dx\) is equal to</p>
<p>(A) \(\frac{x}{10}(5 + 2\sin(2\ln x) + \cos(2\ln x)) + C\)</p>
<p>(B) \(\frac{x}{10}(5 + 2\sin(2\ln x) - \cos(2\ln x)) + C\)</p>
<p>(C) \(\frac{x}{10}(5 - 2\sin(2\ln x) - \cos(2\ln x)) + C\)</p>
<p>(D) \(\frac{x}{10}(5 - 2\sin(2\ln x) + \cos(2\ln x)) + C\)</p>
Step-by-Step Solution
Key Concept: Use the identity sin²(u) = (1 - cos(2u))/2 to convert the integrand into a more manageable form, then apply integration by parts strategically to handle the resulting terms.
<p><strong>Step 1:</strong> Apply the double angle identity: sin²(ln x) = (1 - cos(2ln x))/2</p><p>∫sin²(ln x)dx = ∫(1 - cos(2ln x))/2 dx = (1/2)∫dx - (1/2)∫cos(2ln x)dx = x/2 - (1/2)∫cos(2ln x)dx</p><p><strong>Step 2:</strong> For ∫cos(2ln x)dx, use integration by parts with u = cos(2ln x), dv = dx</p><p>du = -2sin(2ln x)·(1/x)dx, v = x</p><p>∫cos(2ln x)dx = x·cos(2ln x) + 2∫sin(2ln x)dx</p><p><strong>Step 3:</strong> For ∫sin(2ln x)dx, use integration by parts again with u = sin(2ln x), dv = dx</p><p>du = 2cos(2ln x)·(1/x)dx, v = x</p><p>∫sin(2ln x)dx = x·sin(2ln x) - 2∫cos(2ln x)dx</p><p><strong>Step 4:</strong> Let I = ∫cos(2ln x)dx. From steps 2 and 3:</p><p>I = x·cos(2ln x) + 2[x·sin(2ln x) - 2I]</p><p>I = x·cos(2ln x) + 2x·sin(2ln x) - 4I</p><p>5I = x·cos(2ln x) + 2x·sin(2ln x)</p><p>I = (x/5)[cos(2ln x) + 2sin(2ln x)]</p><p><strong>Step 5:</strong> Substitute back into the original expression:</p><p>∫sin²(ln x)dx = x/2 - (1/2)·(x/5)[cos(2ln x) + 2sin(2ln x)] + C</p><p>= x/2 - (x/10)[cos(2ln x) + 2sin(2ln x)] + C</p><p>= (x/10)[5 - cos(2ln x) - 2sin(2ln x)] + C</p><p>= (x/10)[5 - 2sin(2ln x) + cos(2ln x)] + C</p><p><strong>∴ Answer:</strong> D</p>
Correct Answer: D