Question

这道不定积分用换元积分法怎么做呢？

Answer 1

用第一换元法，即凑微分法可以求出结果。

Answer 2

∫(2sinx-cosx)/[3(sinx)^2+4(cosx)^2] dx
=2∫sinx/[3(sinx)^2+4(cosx)^2] dx -∫cosx/[3(sinx)^2+4(cosx)^2] dx
=2∫sinx/[ (cosx)^2+3] dx -∫cosx/[4-(sinx)^2] dx
=-2∫dcosx/[ (cosx)^2+3] -∫ dsinx/[4-(sinx)^2]
=-(2/3)∫dcosx/[ (cosx/√3)^2+1] -(1/4)∫ dsinx/[1-(sinx/2)^2]
=-(2√3/3)∫d(cosx/√3)/[ (cosx/√3)^2+1] -(1/2)∫ d(sinx/2)/[1-(sinx/2)^2]
=-(2√3/3)arctan(cosx/√3) -(1/2)∫ d(sinx/2)/{ [1-(sinx/2)] [1+(sinx/2)] }
=-(2√3/3)arctan(cosx/√3) -(1/4)∫ { 1/[1-(sinx/2)] +1/[1+(sinx/2)] } d(sinx/2)
=-(2√3/3)arctan(cosx/√3) -(1/4)ln|[1+(sinx/2)]/[1-(sinx/2)]| +C