<p>The value of $$\lim_{x \to 0^+} \frac{\int_1^{\cos x} (\cos^{-1} t) \, dt}{2x - \sin 2x}$$ is equal to:</p>
Step-by-Step Solution
Key Concept: This is a 0/0 indeterminate form requiring L'Hôpital's rule. We need to differentiate both numerator and denominator, using Leibniz integral rule for the numerator's derivative.
<p><strong>Step 1: Verify indeterminate form</strong></p><p>At x → 0⁺: Numerator = ∫₁^1 (cos⁻¹ t) dt = 0</p><p>Denominator = 2(0) - sin(0) = 0</p><p>This is 0/0 form, so we apply L'Hôpital's rule.</p><p><strong>Step 2: Differentiate numerator using Leibniz rule</strong></p><p>If N(x) = ∫₁^(cos x) (cos⁻¹ t) dt, then:</p><p>dN/dx = cos⁻¹(cos x) · d(cos x)/dx = cos⁻¹(cos x) · (-sin x)</p><p>For x ∈ (0, π), we have cos⁻¹(cos x) = x, so:</p><p>dN/dx = x · (-sin x) = -x sin x</p><p><strong>Step 3: Differentiate denominator</strong></p><p>If D(x) = 2x - sin 2x, then:</p><p>dD/dx = 2 - 2cos 2x</p><p><strong>Step 4: Apply L'Hôpital's rule</strong></p><p>lim(x → 0⁺) [dN/dx]/[dD/dx] = lim(x → 0⁺) [-x sin x]/[2 - 2cos 2x]</p><p>At x = 0: Numerator = 0, Denominator = 2 - 2(1) = 0</p><p>Still 0/0, apply L'Hôpital's again.</p><p><strong>Step 5: Second application of L'Hôpital's rule</strong></p><p>d(-x sin x)/dx = -sin x - x cos x</p><p>d(2 - 2cos 2x)/dx = 4 sin 2x</p><p>lim(x → 0⁺) [-sin x - x cos x]/[4 sin 2x]</p><p>At x = 0: Numerator = 0, Denominator = 0. Apply L'Hôpital's once more.</p><p><strong>Step 6: Third application of L'Hôpital's rule</strong></p><p>d(-sin x - x cos x)/dx = -cos x - cos x + x sin x = -2cos x + x sin x</p><p>d(4 sin 2x)/dx = 8 cos 2x</p><p>lim(x → 0⁺) [-2cos x + x sin x]/[8 cos 2x]</p><p><strong>Step 7: Evaluate the limit</strong></p><p>= [-2cos(0) + 0·sin(0)]/[8 cos(0)]</p><p>= [-2(1) + 0]/[8(1)]</p><p>= -2/8 = -1/4</p><p><strong>∴ Answer: d</strong></p>
Correct Answer: d