计算定积分 ∫[上限e^(π⼀2)下限1]sin(lnx)dx

2025-04-13 08:52:02
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回答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...