∫(lnx-1)dx/x^2∫(下限0,上限π/2)cosxdx/(cosx+sinx)

Question

∫(lnx-1)dx/x^2
∫(下限0,上限π/2)cosxdx/(cosx+sinx)

Answer 1

∫(lnx-1)dx/x^2
=∫(lnx-1)d(-1/x) (分部积分)
=(-1/x)(lnx-1)-∫(-1/x)d(lnx-1)
=(1-lnx)/x+∫1/x^2dx
=-lnx/x+C
记M=∫(下限0,上限π/2)cosxdx/(cosx+sinx)
N=∫(下限0,上限π/2)sinxdx/(cosx+sinx)
则M+N=∫(下限0,上限π/2)(cosx+sinx)dx/(cosx+sinx)
=∫(下限0,上限π/2) 1dx=π/2
M-N=∫(下限0,上限π/2)(cosx-sinx)dx/(cosx+sinx)
=∫(下限0,上限π/2) d(sinx+cosx)/(cosx+sinx)
=ln|sinx+cosx|(π/2)-ln|sinx+cosx|(0)
=0
所以M+N=π/2,M-N=0,M=N=π/4