Question

恳求大神帮帮我解答一下定积分∫ ln(cos x)dx,求过程，谢谢！

恳求大神帮帮我解答一下定积分∫ ln(cos x)dx,求过程，谢谢！数学分析

Answer 1

原式=∫(0,1)ln(x+1)d(x+1)
=(x+1)ln(x+1) (0,1)-∫(0,1)(x+1)dln(x+1)
=(x+1)ln(x+1) (0,1)-∫(0,1)(x+1)*1/ln(x+1) dx
=(x+1)ln(x+1) (0,1)-∫(0,1) dx
=[(x+1)ln(x+1)-x] (0,1)
=2ln2-1