Question

计算定积分:∫cosx(1+sinx)dx,(区间0到π/2 )？

Answer 1

=∫cosxdx+∫sinxcosxdx
=sinx+(1/2)∫sin2xdx
=sin(π/2)-sin0+(1/4)∫sin2xd2x
=1-(1/4)cos2x
=1-(1/4)(cosπ-cos0)
=1+1/2
=1.5,9,原式=(cosx＋1/2sin2x)dx=sinx－1/4cos2x
再代入就行了，结果是3/2,2,∫cosx(1+sinx)dx
=∫(1+sinx)dsinx
=sinx+1/2(sinx)^2
=sin(π/2)-sin(0)+1/2[sin(π/2)]^2-1/2[sinx(0)]^2
=3/2,2,∫cosx(1+sinx)dx,(区间0到π/2 )
=∫(1+sinx)dsinx,(区间0到π/2 )
=(1/2)(1+sinx)^2,(区间0到π/2 )
=(1/2)*2^2-(1/2)
=3/2,1,