Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>Let \( I = \int_{-\pi/2}^{\pi/2} \sin^4 x \left(1 + \log\left(\dfrac{2+\sin x}{2-\sin x}\right)\right) dx \). Find the value of \(I\).</p>
<p>\( \dfrac{3\pi}{8} \)</p>
<p>\( \dfrac{3\pi}{4} \)</p>
<p>\( \dfrac{\pi}{4} \)</p>
<p>\( \dfrac{\pi}{8} \)</p>
Step-by-Step Solution
Key Concept: Split the integral using the property that f(x) = f_even(x) + f_odd(x). The logarithmic term is odd, so its integral over a symmetric interval vanishes, leaving only the even part ∫sin⁴x dx.
<p><strong>Step 1: Identify odd and even components</strong></p><p>Let f(x) = sin⁴x and g(x) = log((2+sin x)/(2-sin x))</p><p>Check if g(x) is odd: g(-x) = log((2-sin x)/(2+sin x)) = -log((2+sin x)/(2-sin x)) = -g(x) ✓</p><p><strong>Step 2: Split the integral</strong></p><p>I = ∫₋π/₂^π/₂ sin⁴x·log((2+sin x)/(2-sin x)) dx + ∫₋π/₂^π/₂ sin⁴x dx</p><p><strong>Step 3: Apply odd function property</strong></p><p>Since sin⁴x is even and log((2+sin x)/(2-sin x)) is odd, their product is odd.</p><p>∫₋π/₂^π/₂ [even × odd] dx = 0</p><p>Therefore: I = ∫₋π/₂^π/₂ sin⁴x dx</p><p><strong>Step 4: Use reduction formula for ∫sin⁴x dx</strong></p><p>sin⁴x = (3 - 4cos(2x) + cos(4x))/8</p><p>∫₋π/₂^π/₂ sin⁴x dx = ∫₋π/₂^π/₂ (3 - 4cos(2x) + cos(4x))/8 dx</p><p>= (1/8)[3x - 2sin(2x) + sin(4x)/4]₋π/₂^π/₂</p><p>= (1/8)[3π - 0 + 0 - (-3π - 0 - 0)]</p><p>= (1/8)(6π) = <strong>3π/4</strong></p><p>∴ I = <strong>3π/4</strong></p>
Correct Answer: A