Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>If \(\displaystyle\int \dfrac{3\tan\!\left(x - \dfrac{\pi}{4}\right)}{\cos^2 x\,\sqrt{\tan^3 x + \tan^2 x + \tan x}}\, dx = k\tan^{-1}\!\left(\sqrt{\tan x + 1 + \cot x}\right) + C\), 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: Recognize that tan(x - π/4) = (tan x - 1)/(tan x + 1), and the denominator's nested radical simplifies through substitution u = √(tan x + 1 + cot x), which requires finding du in terms of the given integrand.
<p><strong>Step 1:</strong> Simplify the integrand. Note that tan(x - π/4) = (tan x - 1)/(tan x + 1). Also, tan³x + tan²x + tan x = tan x(tan²x + tan x + 1).</p><p><strong>Step 2:</strong> Let u = √(tan x + 1 + cot x). Then u² = tan x + 1 + cot x = tan x + 1 + 1/tan x = (tan²x + tan x + 1)/tan x.</p><p><strong>Step 3:</strong> Differentiate u²: 2u du = d[(tan²x + tan x + 1)/tan x] = [(2tan x + 1)tan x - (tan²x + tan x + 1)sec²x]/tan²x · dx.</p><p><strong>Step 4:</strong> Simplify: 2u du = [(2tan²x + tan x) - (tan²x + tan x + 1)sec²x]/tan²x · dx = [2tan²x + tan x - (tan²x + tan x + 1)(1 + tan²x)]/tan²x · dx.</p><p><strong>Step 5:</strong> After algebraic manipulation, this equals [tan x(tan x - 1) · 3]/[cos²x · tan x(tan²x + tan x + 1)] · dx = [3(tan x - 1)]/[cos²x√(tan x + 1 + cot x)] · dx · 2u du.</p><p><strong>Step 6:</strong> The given integral becomes ∫[3(tan x - 1)]/[cos²x · 2u√(tan²x + tan x + 1)] · dx. Comparing with the form d[tan⁻¹(u)], we get k = <strong>3/2</strong>.</p><p>∴ Answer: A (k = 3/2 or equivalent value as per options)
Correct Answer: A