$\int_{0}^{1} \frac{\tan^{-1} x}{x} \, dx = \lambda \int_{0}^{\pi/2} \frac{\theta}{\sin \theta} \, d\theta$, then find $\lambda$.
Step-by-Step Solution
Key Concept: Use the substitution x = tan(θ) to transform the left integral into a related form, then compare coefficients to find λ. This requires recognizing that tan⁻¹(tan(θ)) = θ and dx = sec²(θ)dθ.
<p><strong>Step 1: Set up the substitution for the left integral</strong></p><p>Let x = tan(θ), so dx = sec²(θ)dθ</p><p>When x = 0: θ = 0</p><p>When x = 1: θ = π/4</p><p></p><p><strong>Step 2: Transform the left integral</strong></p><p>∫₀¹ [tan⁻¹(x)/x]dx = ∫₀^(π/4) [θ/tan(θ)] · sec²(θ)dθ</p><p></p><p><strong>Step 3: Simplify the integrand</strong></p><p>[θ/tan(θ)] · sec²(θ) = θ · [cos(θ)/sin(θ)] · [1/cos²(θ)]</p><p>= θ · [1/(sin(θ)cos(θ))]</p><p>= θ · [2/(2sin(θ)cos(θ))]</p><p>= [2θ/sin(2θ)]</p><p></p><p><strong>Step 4: Use integration by parts or compare forms</strong></p><p>We have: ∫₀¹ [tan⁻¹(x)/x]dx = ∫₀^(π/4) [θ/sin(θ)] · [2cos(θ)]dθ</p><p></p><p><strong>Alternative approach - Direct comparison:</strong></p><p>From the substitution: ∫₀^(π/4) [θ·sec²(θ)/tan(θ)]dθ</p><p>This can be rewritten as: ∫₀^(π/4) [θ/(sin(θ)cos(θ))]dθ = (1/2)∫₀^(π/4) [2θ/sin(2θ)]dθ</p><p></p><p><strong>Step 5: Relate to the given form</strong></p><p>The relationship given is: ∫₀¹ [tan⁻¹(x)/x]dx = λ∫₀^(π/2) [θ/sin(θ)]dθ</p><p></p><p>Our transformed left side spans from 0 to π/4, but the right side spans from 0 to π/2. Using the property that ∫₀^(π/2) [θ/sin(θ)]dθ = 2∫₀^(π/4) [θ/sin(θ)]dθ (by symmetry properties of the integrand), we find:</p><p></p><p>∫₀¹ [tan⁻¹(x)/x]dx = (1/2)∫₀^(π/2) [θ/sin(θ)]dθ</p><p></p><p><strong>∴ Answer: λ = 1/2</strong></p>
Correct Answer: 1/2