Definite Integration Question - None

Question

The value of the integral $\int_{-\pi/2}^{\pi/2} \sin^4 x \left(1 + \log\left(\dfrac{2+\sin x}{2-\sin x}\right)\right) dx$ is

Accepted Answer

Step 1: Split the integral into two parts:

∫_-π/2^π/2 sin⁴x(1 + log((2+sin x)/(2-sin x))) dx = ∫_-π/2^π/2 sin⁴x dx + ∫_-π/2^π/2 sin⁴x · log((2+sin x)/(2-sin x)) dx

Step 2: Analyze the second integral. Let f(x) = sin⁴x · log((2+sin x)/(2-sin x))

Check: f(-x) = sin⁴(-x) · log((2+sin(-x))/(2-sin(-x))) = sin⁴x · log((2-sin x)/(2+sin x)) = -sin⁴x · log((2+sin x)/(2-sin x)) = -f(x)

So the second integral is an odd function over symmetric limits, hence equals 0.

Step 3: Evaluate ∫_-π/2^π/2 sin⁴x dx using the reduction formula or identity:

sin⁴x = (3 - 4cos 2x + cos 4x)/8

∫_-π/2^π/2 sin⁴x dx = ∫_-π/2^π/2 (3 - 4cos 2x + cos 4x)/8 dx

= (1/8)[3x - 2sin 2x + (sin 4x)/4]_-π/2^π/2

= (1/8)[3(π/2 + π/2) - 0] = (1/8) · 3π = 3π/8

∴ Answer: D

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