Question Detail
Question
Given that ∫0θ 4sin(2x)cos⁴(2x) dx = 4/5, where 0 < θ < π. Find θ.
Correct Answer
Option A is the correct answer.
Detailed Explanation
Let u = cos(2x).Then du/dx = -2sin(2x), so du = -2sin(2x) dx, or sin(2x) dx = -du/2.The integral becomes ∫4u⁴(-du/2) = -2∫u⁴ du = -2(u⁵/5) + C = -2cos⁵(2x)/5 + C.Now, evaluate the definite integral from 0 to θ:[-2cos⁵(2x)/5]0θ = (-2cos⁵(2θ)/5) - (-2cos⁵(0)/5)= -2cos⁵(2θ)/5 + 2(1)⁵/5 = -2cos⁵(2θ)/5 + 2/5.We are given that this equals 4/5:-2cos⁵(2θ)/5 + 2/5 = 4/5-2cos⁵(2θ)/5 = 2/5-2cos⁵(2θ) = 2cos⁵(2θ) = -1cos(2θ) = -1.Since 0 < θ < π, we have 0 < 2θ < 2π.The only value in this interval for which cos(2θ) = -1 is 2θ = π.Therefore, θ = π/2.
Hint
Use u-substitution with u = cos(2x). Carefully handle the limits of integration or substitute back before evaluating.