<p>If \(\int \frac{dx}{\cos^3 x \cdot \sqrt{2\sin 2x}} = (\tan x)^A + C(\tan x)^B + k\), where \(k\) is a constant of integration, then \(A + B + C\) equals</p>
Step-by-Step Solution
Key Concept: Rewrite √(2sin 2x) = √(4sin x cos x) = 2√(sin x cos x), then express the integrand in terms of tan x and sec x using the substitution t = tan x (dt = sec²x dx). This converts a trigonometric integral into a polynomial in t.
<p><strong>Step 1:</strong> Simplify the denominator.</p><p>2sin 2x = 4sin x cos x, so √(2sin 2x) = 2√(sin x cos x)</p><p><strong>Step 2:</strong> Rewrite the integral in terms of sin x and cos x.</p><p>∫ dx/(cos³x · 2√(sin x cos x)) = (1/2)∫ dx/(cos³x √sin x √cos x) = (1/2)∫ dx/(cos^(7/2) x √sin x)</p><p><strong>Step 3:</strong> Express in terms of tan x and sec x.</p><p>√sin x = √sin x, and divide numerator and denominator by cos^(7/2) x:</p><p>= (1/2)∫ sec^(7/2) x/√sin x dx</p><p>Rewrite as: (1/2)∫ sec²x · sec^(3/2) x/(√sin x) dx = (1/2)∫ (1 + tan²x)^(1/2) · sec²x/√sin x dx</p><p><strong>Step 4:</strong> Use substitution t = tan x, so dt = sec²x dx and √sin x = t/√(1+t²).</p><p>= (1/2)∫ √(1+t²) · √(1+t²)/|t| dt = (1/2)∫ (1+t²)/|t| dt</p><p><strong>Step 5:</strong> For t > 0: (1/2)∫ (1/t + t) dt = (1/2)[ln|t| + t²/2] = (1/2)ln|tan x| + (1/4)tan²x</p><p>This gives the form: (1/4)(tan x)² + (1/2)ln(tan x) = (tan x)^2/4 + (1/2)(tan x)^0 · ln(tan x)</p><p><strong>Step 6:</strong> Comparing with (tan x)^A + C(tan x)^B, we have A = 2, B = 0, C = 1/2.</p><p>∴ A + B + C = 2 + 0 + 1/2 = <strong>2.5 or 5/2</strong></p>
Correct Answer: B