Question

计算定积分∫[上限e^(π/2)下限1]sin(lnx)dx？

Answer 1

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.,10,