Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>The integral \(\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}\) is equal to (where \(C\) is a constant of integration)</p>
<p>\(-2\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}+C\)</p>
<p>\(-\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C\)</p>
<p>\(-2\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C\)</p>
<p>\(2\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}+C\)</p>
Step-by-Step Solution
Key Concept: Recognize that √(x-x²) = √[x(1-x)] and use the substitution √x = sin θ to convert the denominator into a standard trigonometric form. This transforms the radical expression into something manageable.
<p><strong>Step 1:</strong> Rewrite the integrand by factoring: √(x-x²) = √[x(1-x)] = √x·√(1-x)</p><p><strong>Step 2:</strong> Let √x = sin θ, so x = sin²θ and dx = 2sin θ cos θ dθ. Then √(1-x) = cos θ</p><p><strong>Step 3:</strong> Substitute into the integral:</p><p>∫ (2sin θ cos θ dθ)/[(1+sin θ)·sin θ·cos θ] = ∫ (2 dθ)/(1+sin θ)</p><p><strong>Step 4:</strong> Rationalize by multiplying by (1-sin θ)/(1-sin θ):</p><p>∫ (2(1-sin θ))/(cos²θ) dθ = ∫ (2 sec²θ - 2 sec θ tan θ) dθ</p><p><strong>Step 5:</strong> Integrate: 2 tan θ - 2 sec θ + C</p><p><strong>Step 6:</strong> Convert back using tan θ = √x/√(1-x) and sec θ = 1/√(1-x):</p><p>2·(√x/√(1-x)) - 2/√(1-x) + C = (2√x - 2)/√(1-x) + C = 2(√x - 1)/√(1-x) + C</p><p>∴ Answer: A</p>
Correct Answer: A