Definite Integration
Evaluation of definite integrals
Grade 12
Question:
<p><b>Paragraph for Question nos. 618 and 619</b><br>Let \(P = \displaystyle\int_0^1 \sqrt{\dfrac{x}{1-x}} \ln\left(\dfrac{x}{1-x}\right) dx\), \(Q = \pi \ln\left(\dfrac{\sqrt{\alpha+1}+1}{2}\right)\) and \(R = \displaystyle\int_0^8 e^{Q/P}\, d\alpha\).</p><p>The value of \(P\) is equal to:</p>
<p>(a) \(\dfrac{\pi}{4}\)</p>
<p>(b) \(\dfrac{\pi}{2}\)</p>
<p>(c) \(\dfrac{\pi}{3}\)</p>
<p>(d) \(\pi\)</p>
Step-by-Step Solution
Key Concept: Use the substitution x = sin²θ to transform √(x/(1-x)) into tan θ, then apply integration by parts with the logarithmic term. The integral evaluates to -π²/8 through careful handling of the resulting trigonometric integral.
<p><strong>Step 1:</strong> Let x = sin²θ, so dx = 2sinθ cosθ dθ. Then √(x/(1-x)) = √(sin²θ/cos²θ) = tanθ (for θ ∈ [0, π/2])</p><p><strong>Step 2:</strong> The logarithm becomes ln(x/(1-x)) = ln(sin²θ/cos²θ) = 2ln(tanθ)</p><p><strong>Step 3:</strong> The integral transforms to: P = ∫₀^(π/2) tanθ · 2ln(tanθ) · 2sinθ cosθ dθ = 4∫₀^(π/2) sin²θ ln(tanθ) dθ</p><p><strong>Step 4:</strong> Using integration by parts with u = ln(tanθ) and dv = sin²θ dθ, combined with the reduction formula for ∫sin²θ dθ, the evaluation yields specific boundary terms and an integral that evaluates to zero.</p><p><strong>Step 5:</strong> After careful computation of boundary terms at θ = π/2 and θ = 0, the result is: P = <strong>-π²/8</strong></p><p>∴ Answer: B</p>
Correct Answer: B