计算定积分∫[上限e^(π/2)下限1]sin(lnx)dx
1个回答

设lnx=t,则当x=1时,t-0.当x=e^(π/2)时,t=π/2

∴原式=∫(0,π/2)e^tsintdt （∫(0,π/2)表示从0到π/2积分）

为了求解方便,设I=∫(0,π/2)e^tsintdt

∵I=(e^tsintdt)|(0,π/2)-∫(0,π/2)e^tcostdt (应用分部积分)

==>I=e^(π/2)-(e^tcost)|(0,π/2)-∫(0,π/2)e^tcostdt (应用分部积分)

==>I=e^(π/2)+1-I (∵I=∫(0,π/2)e^tsintdt)

==>2I=e^(π/2)+1

==>I=[e^(π/2)+1]/2

∴原式=[e^(π/2)+1]/2.

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