Question Detail
Question
Given that ∫0θ 4 sin(2x)cos4(2x) dx = 4/5, where 0 < θ < π, find the exact value of θ.
Correct Answer
Option A is the correct answer.
Detailed Explanation
Let u = cos(2x), then du = -2 sin(2x) dx. So, sin(2x) dx = -1/2 du.Substitute into the integral: ∫ 4 * u4 * (-1/2) du = -2 ∫ u4 du = -2 * (u5/5) = -(2/5)cos5(2x).Now evaluate the definite integral from 0 to θ:[-(2/5)cos5(2x)]0θ = -(2/5)[cos5(2θ) - cos5(2*0)]= -(2/5)[cos5(2θ) - cos5(0)]= -(2/5)[cos5(2θ) - 1].We are given that this equals 4/5:-(2/5)[cos5(2θ) - 1] = 4/5.Divide both sides by -2/5:cos5(2θ) - 1 = -2.cos5(2θ) = -1.This implies cos(2θ) = -1.For cos(y) = -1, the general solution is y = (2n+1)π for integer n.So, 2θ = π (since 0 < θ < π, so 0 < 2θ < 2π).θ = π/2.
Hint
Use u-substitution with u = cos(2x) to solve the integral. Then, solve the resulting trigonometric equation for θ within the given range.