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求∫(0-π/2)e^2xcosxdx,∫(0-π)(xsinx)^2dx,∫(π/3-π/4)x/sinx^2dx,∫(0-π^2/4)sin(x^1/2)定积分
人气:115 ℃ 时间:2020-05-08 10:07:05
解答
∫(0-π/2)e^(2x)cosxdx=∫(0-π/2)e^(2x)dsinx=e^(2x)sinx|(0-π/2)-∫(0-π/2)sinxde^(2x)=e^π·sin(π/2)-0-2∫(0-π/2)e^(2x)sinxdx=e^π+2∫(0-π/2)e^(2x)dcosx=e^π+2[e^(2x)cosx|(0-π/2)-∫(0-...
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