<p>∫ <sup>dx</sup>/<sub>√(1-tan²x)</sub> = λ sin⁻¹(λ sinx) + <i>C</i>, then λ = ?</p>
Step-by-Step Solution
Key Concept: Simplify √(1-tan²x) using trigonometric identities, then use substitution to match the given form and determine λ by comparing coefficients.
<p><strong>Step 1:</strong> Simplify the integrand using the identity 1 - tan²x.</p><p>1 - tan²x = 1 - sin²x/cos²x = (cos²x - sin²x)/cos²x = cos(2x)/cos²x</p><p>Therefore: √(1-tan²x) = √[cos(2x)/cos²x] = √[cos(2x)]/|cos x|</p><p><strong>Step 2:</strong> For valid regions where cos(2x) > 0 and cos x > 0:</p><p>∫ dx/√(1-tan²x) = ∫ cos x/√[cos(2x)] dx</p><p><strong>Step 3:</strong> Use the identity cos(2x) = 1 - 2sin²x:</p><p>∫ cos x/√[1-2sin²x] dx</p><p><strong>Step 4:</strong> Apply substitution u = sin x, du = cos x dx:</p><p>∫ du/√[1-2u²] = ∫ du/√[1-(√2 u)²]</p><p><strong>Step 5:</strong> Let v = √2 u, then dv = √2 du:</p><p>∫ (1/√2) dv/√[1-v²] = (1/√2) sin⁻¹(v) + C = (1/√2) sin⁻¹(√2 sin x) + C</p><p><strong>Step 6:</strong> Comparing with λ sin⁻¹(λ sin x) + C:</p><p>λ = 1/√2 · √2 = √2</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c