Question

求定积分:∫ln(1+tanx)dx (o≤x≤π/4)

求不定积分:∫ln(1+tanx)dx (o≤x≤π/4)

Answer 1

∫[0,π/4]ln(1+tanx)dx
换元π/4-t=x
=-∫[π/4,0]ln[1+(1-tant)/(tant+1)]dt=
=∫[0,π/4]ln[2/(tant+1)]dt=∫[0,π/4]ln2-∫[0,π/4]ln(tant+1)dt=πln2/4-∫[0,π/4]ln(tanx+1)dx
2∫[0,π/4]ln(1+tanx)dx=πln2/4
所以∫[0,π/4]ln(1+tanx)dx=πln2/8

Answer 2

换元 令t=x-π/4，然后把tan（t-π/4）打开就很好求了，不信可以自己试着做做

Answer 3

1/1+tanx