Question

我想请教一道定积分的题目：如下,求［0,π/2］上,［（sinx）^10-（cos）^10］/（1+sinx+cosx）

Answer 1

∫[0,π/2](sinx^10-cosx^10)dx/(1+sinx+cosx)

=∫[0,π/4](sinx^10-cosx^10)dx/(1+sinx+cosx)+∫[π/4,π/2](sinx^10-cosx^10)dx/(1+sinx+cosx)

u=π/2-x

=∫[0,π/4](sinx^10-cosx^10)dx/(1+sinx+cosx)+ ∫[π/4,0] (cosu^10-sinu^10)d(-u)/(1+sinu+cosu)

=∫[0,π/4](sinx^10-cosx^10)dx/(1+sinx+cosx)+∫[π/4,0](sinx^10-cosx^10)dx/(1+sinx+cosx)

=∫[0,0](sinx^10-cosx^10)dx/(1+sinx+cosx)

=0