Question

求大神解答一个换元法求积分的问题

请问这一步是怎么来的，谢谢！！！
就是最后一行那个式子

Answer 1

(1) ∫(ln2,ln3)1/[e^x-e^(-x)]dx
=∫(ln2,ln3) e^x/[(e^x)^2-1]dx
=∫(ln2,ln3) 1/[(e^x)^2-1]de^x
令t=e^x
t1=e^ln2=2
t2=e^ln3=3
上式=∫(2,3)1/(t^2-1)dt
=1/2∫(2,3)[1/(t-1)-1/(t+1)]dt
=1/2ln[(t-1)/(t+1)]|(2,3)
=1/2{ln[(3-1)/(3+1)]-ln[(2-1)/(2+1)]}
=1/2[ln(1/2)-ln(1/3)]
=1/2ln[(1/2)/(1/3)]
=1/2ln(3/2)
(2) ∫(-π/2,π/2) √(cosx-cos^3x)dx