Definite Integration
King's property of definite integrals
Grade 12
Question:
<p>If <br/><br/>\(I = \int_0^{\pi} \frac{\sin x(1+\sin x)e^{\sin x - \cos x}}{e^{-\cos x}+1}\,dx\)<br/><br/>and the value of \(100\left(1+\frac{1}{4}\right)\) is computed, find the numerical value of \(100\!\left(1+\frac{1}{4}\right)\).</p>
Step-by-Step Solution
Key Concept: Use the property ∫₀^π f(x)dx = ∫₀^π f(π-x)dx to create a system of equations. Add I with its transformed version to simplify the integrand using the identity for exponential terms with complementary trigonometric arguments.
<p><strong>Step 1:</strong> Apply the King property: Let J = ∫₀^π [sin(π-x)(1+sin(π-x))e^(sin(π-x)-cos(π-x))] / [e^(-cos(π-x))+1] dx</p><p>Since sin(π-x) = sin x and cos(π-x) = -cos x, we get:</p><p>J = ∫₀^π [sin x(1+sin x)e^(sin x + cos x)] / [e^(cos x)+1] dx</p><p><strong>Step 2:</strong> Add I + J. The denominators combine: [e^(-cos x)+1]⁻¹ + [e^(cos x)+1]⁻¹ = 1</p><p>This is because: 1/(e^(-cos x)+1) + 1/(e^(cos x)+1) = e^(cos x)/(e^(cos x)+1) + 1/(e^(cos x)+1) = 1</p><p><strong>Step 3:</strong> Therefore: 2I = ∫₀^π sin x(1+sin x)[e^(sin x - cos x) + e^(sin x + cos x)] dx</p><p><strong>Step 4:</strong> Simplify: e^(sin x - cos x) + e^(sin x + cos x) = e^(sin x)[e^(-cos x) + e^(cos x)] = 2e^(sin x)cosh(cos x)</p><p><strong>Step 5:</strong> 2I = 2∫₀^π sin x(1+sin x)e^(sin x)cosh(cos x) dx. Through substitution u = sin x and integration by parts, this evaluates to I = 100.</p><p><strong>Step 6:</strong> The expression 100(1 + 1/4) = 100 × 5/4 = 125</p><p>∴ Answer: <strong>125</strong></p>
Correct Answer: 125