[sin(lnx)]^2的不定积分怎么求啊?
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[sin(lnx)]^2的不定积分怎么求啊?
![[sin(lnx)]^2的不定积分怎么求啊?](/uploads/image/z/18905815-55-5.jpg?t=%5Bsin%28lnx%29%5D%5E2%E7%9A%84%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E6%80%8E%E4%B9%88%E6%B1%82%E5%95%8A%3F)
∫[sin(lnx)]^2dx
=(1/2)∫(1-cos(2lnx))dx
=x/2-(1/2)∫cos(2lnx)dx
∫cos(2lnx)dx=xcos(2lnx)-∫xdcos(2lnx)=xcos(2lnx)+2∫sin(2lnx)dx
=xcos(2lnx)+2xsin(2lnx)-4∫cos(2lnx)dx
5∫cos(2lnx)dx=xcos(2lnx)+2xsin(2lnx)
∫cos(2lnx)dx=(1/5)xcos(2lnx) +(2/5)xsin(2lnx)
=(1/2)∫(1-cos(2lnx))dx
=x/2-(1/2)∫cos(2lnx)dx
∫cos(2lnx)dx=xcos(2lnx)-∫xdcos(2lnx)=xcos(2lnx)+2∫sin(2lnx)dx
=xcos(2lnx)+2xsin(2lnx)-4∫cos(2lnx)dx
5∫cos(2lnx)dx=xcos(2lnx)+2xsin(2lnx)
∫cos(2lnx)dx=(1/5)xcos(2lnx) +(2/5)xsin(2lnx)