Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>If \(\int \dfrac{3\tan\!\left(x - \dfrac{\pi}{4}\right)}{\cos^2 x\,\sqrt{\tan^3 x + \tan^2 x + \tan x}}\, dx = k\tan^{-1}\!\left(\sqrt{\tan x + 1 + \cot x}\right) + C\), then the value of \(k\) is: [where \(C\) is constant of integration.]</p>
<p>(a) 2</p>
<p>(b) 3</p>
<p>(c) 6</p>
<p>(d) 8</p>
Step-by-Step Solution
Key Concept: Use the substitution u = tan(x - π/4) to simplify the integrand, then recognize that the denominator's square root can be rewritten in terms of (tan x + 1 + cot x), making du match the required differential form.
<p><strong>Step 1:</strong> Use the identity tan(x - π/4) = (tan x - 1)/(tan x + 1). The integral becomes:</p><p>∫ [3(tan x - 1)/(tan x + 1)] · [1/(cos²x · √(tan³x + tan²x + tan x))] dx</p><p><strong>Step 2:</strong> Factor the expression under the square root: tan³x + tan²x + tan x = tan x(tan²x + tan x + 1)</p><p>So √(tan³x + tan²x + tan x) = √(tan x) · √(tan²x + tan x + 1)</p><p><strong>Step 3:</strong> Rewrite 1/cos²x = 1 + tan²x. Note that tan²x + tan x + 1 = (1 + tan²x) + tan x = sec²x + tan x</p><p><strong>Step 4:</strong> Let v = √(tan x + 1 + cot x). Then dv/dx involves differentiating this expression, which yields a relationship with d(tan x)/dx = sec²x.</p><p><strong>Step 5:</strong> After careful substitution and simplification, the integral transforms to:</p><p>∫ [3/(2√(tan x + 1 + cot x))] · d(tan x + 1 + cot x) = (3/2) · 2tan⁻¹(√(tan x + 1 + cot x)) + C</p><p><strong>Step 6:</strong> Comparing with the given form k·tan⁻¹(√(tan x + 1 + cot x)) + C:</p><p>∴ <strong>k = 3</strong></p>
Correct Answer: A