Question

用换元法计算定积分，急求大神解答

Answer 1

追问

大神，顺便

追答

（6）分部积分d（-cotx）
（7）=[xln（1+x²）]_0^1-2∫（0,1）x²/（1+x²）dx

Answer 2

(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
=∫(-π/2,π/2) √[cosx(1-cos^2x)]dx
=∫(-π/2,π/2) |sinx|√cosxdx
∵|sinx|√cosx是偶函数
∴上式=2∫(0,π/2) sinx√cosxdx
=-2∫(0,π/2)√cosxdcosx
令t=cosx
t1=cos0=1
t2=cosπ/2=0
上式=-2∫(1,0)√tdt
=-2×2/3t^(3/2)|(1,0)
=-4/3[0^(3/2)-1^(3/2)]
=-4/3[0-1]
=4/3