求定积分∫(1+sinx)^2dx从-π/2到π/2

Question

求定积分∫(1+sinx)^2dx 从-π/2到π/2

Answer 1

∫(1+sinx)^2dx 从-π/2到π/2
= ∫(- π/2→π/2) (1 + 2sinx + sin²x) dx
= ∫(- π/2→π/2) (1 + sin²x) dx + 2∫(- π/2→π/2) sinx dx
= 2∫(0→π/2) (1 + sin²x) dx + 0
= 2∫(0→π/2) [1 + (1 - cos(2x))/2] dx
= 2∫(0→π/2) [3/2 - (1/2)cos(2x)] dx
= 2[3x/2 - (1/4)sin(2x)] |(0→π/2)
= 2[3/2 * π/2 - 0]
= 3π/2