Evaluate the interval

Question

∫_-π^π(3cos(Θ-π/3))dΘ

Answer 1

Let u = θ - π/3. Then dθ = du

When θ = -π, u = -4π/3 and when θ = π, u = 2π/3

∫3cos(θ-π/3)dθ = 3∫cosudu = 3sinu + C

So, the value of the given definite integral is 3[sin(2π/3) - sin(-4π/3)] = 3[√3/2 - √3/2] = 0