∫π⼀2到0 sinX⼀(sinX+cosX) dX<<<这个定积分怎么算??

2024-12-14 05:52:52
推荐回答(1个)
回答1:

∫(π/2->0) sinx/(sinx+cosx) dx
= (1/2)∫(π/2->0) 2sinx/(sinx+cosx) dx
= (1/2)∫(π/2->0) [(sinx+cosx)+(sinx-cosx)]/(sinx+cosx) dx
= (1/2)∫(π/2->0) dx + (1/2)∫(π/2->0) (sinx-cosx)/(sinx+cosx) dx
= (1/2)(-π/2) - (1/2)∫(π/2->0) d(cosx+sinx)/(sinx+cosx)
= -π/4 - (1/2)ln|sinx+cosx|
= -π/4 - (1/2)[ln(1) - ln(1)]
= -π/4