Question

求定积分：∫【π/2 0】sinx/(3+cosx)^2 dx

Answer 1

(cosx)'=-sinx所以：

∫[π/2 0]sinx/(3+cosx)^2 dx

=-∫[π/2 0]1/(3+cosx)^2d(cosx)

=[1/(3+cosx)][π/2 0]

=1/3-1/4

=1/12