Question

定积分问题

第三小题求解

Answer 1

如下

Answer 2

（3） 原积分 I = ∫<0, π/4>xd(1+tanx)/(1+tanx)^2
= -∫<0, π/4>xd[1/(1+tanx)]
= -[x/(1+tanx)]<0, π/4> + ∫<0, π/4>dx/(1+tanx),
= -π/8 + ∫<0, π/4>cosxdx/(cosx+sinx)
= -π/8 + ∫<0, π/4>cos(x-π/4+π/4)dx/[√2cos(x-π/4)]
= -π/8 + ∫<0, π/4>(√2/2)[cos(x-π/4)-sin(x-π/4)]dx/[√2cos(x-π/4)]
= -π/8 + (1/2)∫<0, π/4>[1-tan(x-π/4)]dx
= -π/8 + π/8 - (1/2)∫<0, π/4>tan(x-π/4)d(x-π/4)
= (1/2)[lncos(x-π/4)]<0, π/4> = (1/4)ln2

Answer 3

在积分学中，只要积分区间相同，被积函数式也相同，那么无论自变量用什么字母表示，积分值都不变。