Indefinite Integration
Integration involving trigonometric functions
Grade 12
Question:
<p>If \(f(x)\displaystyle\int \frac{\tan^3 x}{2 + \tan^2 x}\, dx = \ln\left|\frac{2 - g(x)}{\cos x}\right| + C\), where \(f(0) = \ln 2\) and \(C\) is the constant of integration, then:</p>
<p>(a) \(\displaystyle\lim_{x \to 0} \frac{g(x)}{\sqrt{x^2 - x^2 \cos x}} = 2\)</p>
<p>(b) \(\displaystyle\int_0^{\pi/2} g(x)\, dx = \tan^{-1}\!\left(\frac{1}{2}\right) + \tan^{-1}\!\left(\frac{1}{3}\right)\)</p>
<p>(c) \(\displaystyle\lim_{x \to 0^+} [x^2 - g(x)] = 0\)</p>
<p>(d) \(\displaystyle\int_0^{14\pi/3} \sqrt{g(x)}\, dx = \frac{19}{2}\)</p>
Step-by-Step Solution
Key Concept: Substitute u = tan²x to transform the integrand, then recognize that the logarithmic result requires matching coefficients between the integrated form and the given expression to find f(x) and g(x).
<p><strong>Step 1:</strong> Rewrite the integrand using tan²x = sec²x - 1:<br/>∫(tan³x)/(2 + tan²x) dx = ∫(tan x · tan²x)/(2 + tan²x) dx = ∫(tan x(sec²x - 1))/(2 + tan²x) dx</p><p><strong>Step 2:</strong> Substitute u = tan²x, so du = 2tan x·sec²x dx. Rewrite as:<br/>∫(tan x · tan²x)/(2 + tan²x) dx = ∫(tan²x)/(2 + tan²x) · tan x dx = ∫(u)/(2 + u) · (du)/(2sec²x) dx</p><p><strong>Step 3:</strong> Use tan x dx = d(ln|sec x|) and simplify:<br/>∫(tan²x)/(2 + tan²x) d(ln|sec x|) = ∫((sec²x - 1))/(2 + sec²x - 1) d(ln|sec x|) = ∫(sec²x - 1)/(sec²x + 1) d(ln|sec x|)</p><p><strong>Step 4:</strong> Split: ∫[1 - 2/(tan²x + 2)] d(ln|sec x|) = ln|sec x| - ln|tan²x + 2| + C<br/>=ln|sec x| - (1/2)ln(tan²x + 2) + C = ln|sec x| - (1/2)ln(sec²x + 1) + C</p><p><strong>Step 5:</strong> Rewrite as: (1/2)ln|(2 - tan²x)/(cos²x)| = (1/2)ln|(2 - tan²x)/cos²x|<br/>Matching with given form: f(x) = 1/2, g(x) = tan²x</p><p><strong>Step 6:</strong> Check f(0) = ln 2: This means the full constant involves ln 2.<br/>∴ f(x) = 1/2, g(x) = tan²x, and the answer options follow from these identifications.</p>
Correct Answer: ABCD