<p><strong>26.</strong> If the value of definite integral \(\displaystyle\int_{\pi/4}^{\pi/3} e^x \left(\dfrac{2 + \sin 2x}{1 + \cos 2x}\right) dx\) is expressed as \(e^{\frac{\pi}{a}}\left(be^{\frac{\pi}{c}} - 1\right)\), then the value of \(\dfrac{b^2 c}{a}\) is:</p>
Step-by-Step Solution
Key Concept: Simplify the integrand by recognizing that (2 + sin 2x)/(1 + cos 2x) = 1 + tan x, then use integration by parts with u = tan x and dv = e^x dx, leveraging the derivative relationship d/dx(tan x) = sec²x.
<p><strong>Step 1: Simplify the integrand</strong></p><p>Note that 1 + cos 2x = 2cos²x and sin 2x = 2sin x cos x</p><p>So: (2 + sin 2x)/(1 + cos 2x) = (2 + 2sin x cos x)/(2cos²x) = (1 + sin x cos x)/cos²x</p><p>This simplifies to: sec²x + tan x = d/dx(tan x) + tan x</p><p><strong>Step 2: Rewrite the integral</strong></p><p>∫ e^x(sec²x + tan x)dx = ∫ e^x·d/dx(tan x)dx + ∫ e^x·tan x dx</p><p><strong>Step 3: Apply integration by parts</strong></p><p>For ∫ e^x·d/dx(tan x)dx, use parts: u = tan x, dv = e^x dx</p><p>This gives: [e^x·tan x] - ∫ e^x·tan x dx + ∫ e^x·tan x dx = e^x·tan x</p><p><strong>Step 4: Evaluate at bounds</strong></p><p>∫_{π/4}^{π/3} e^x(2 + sin 2x)/(1 + cos 2x) dx = [e^x·tan x]_{π/4}^{π/3}</p><p>= e^(π/3)·tan(π/3) - e^(π/4)·tan(π/4)</p><p>= e^(π/3)·√3 - e^(π/4)·1</p><p>= e^(π/4)(e^(π/12)·√3 - 1)</p><p><strong>Step 5: Match with given form</strong></p><p>Expressing as e^(π/a)(b·e^(π/c) - 1):</p><p>a = 4, b = √3, c = 12</p><p>∴ b²c/a = (√3)²·12/4 = 3·12/4 = <strong>9</strong></p>
Correct Answer: D