Definite Integration
Integration by substitution
Grade 12
Question:
<p>Evaluate the definite integral: \[I = \int_0^1 \frac{e^x + e^{-x}}{\sqrt{11 - e^{2x} - e^{-2x}}}\, dx\]</p>
<p>\(\sin^{-1}\left(\dfrac{e - e^{-1}}{3}\right)\)</p>
<p>\(\cos^{-1}\left(\dfrac{e - e^{-1}}{3}\right)\)</p>
<p>\(\tan^{-1}\left(\dfrac{e - e^{-1}}{3}\right)\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
Step-by-Step Solution
Key Concept: Recognize that e^x + e^(-x) is the derivative of e^x - e^(-x), and the denominator √(11 - e^(2x) - e^(-2x)) can be rewritten using the identity e^(2x) + e^(-2x) = (e^x + e^(-x))² - 2. Substitute u = e^x - e^(-x) to transform into a standard inverse sine integral.
<p><strong>Step 1:</strong> Let u = e^x - e^(-x), then du = (e^x + e^(-x))dx</p><p><strong>Step 2:</strong> Rewrite the denominator using e^(2x) + e^(-2x) = (e^x - e^(-x))² + 2:</p><p>11 - e^(2x) - e^(-2x) = 11 - [(e^x - e^(-x))² + 2] = 9 - u²</p><p><strong>Step 3:</strong> When x = 0: u = e^0 - e^0 = 0</p><p>When x = 1: u = e - e^(-1) = e - 1/e = (e² - 1)/e</p><p><strong>Step 4:</strong> Transform the integral:</p><p>I = ∫₀^((e²-1)/e) du/√(9 - u²) = [arcsin(u/3)]₀^((e²-1)/e)</p><p><strong>Step 5:</strong> I = arcsin((e² - 1)/(3e)) - arcsin(0) = arcsin((e² - 1)/(3e))</p><p>∴ Answer: A</p>
Correct Answer: A