在 2*arctan2 到兀/4 上求函数（1－cosX）（sinX）^2的定积分.求详解.

∫(1-cosx)(sinx)^2dx=∫(sinx)^2dx-∫(sinx)^2dsinx
=∫(1/2-cos2x/2)dx-(sinx)^3/3
=x/2-sin2x/4-(sinx)^3/3
又sin2x=2tanx/[1+(tanx)^2]
sin(2arctanx)=2x/(1+x^2)
sin(2arctan2)=4/5
sin4x=2sin2xcos2x=4tanx[1-(tanx)^2]/[1+(tanx)^2]^2
sin(4arctanx)=4x(1-x^2)/(1+x^2)^2
sin(4arctan2)=8(1-4)/(1+4)^2=-24/25
所以
(2*arctan2,π/4)∫(1-cosx)(sinx)^2dx=(2*arctan2,π/4)[x/2-sin2x/4-(sinx)^3/3]
=π/8-1/4-√2/12-arctan2-6/25+64/375
=π/8-arctan2-√2/12-349/375