Question

一个数学分析问题

ln ((b的平方)(cosx的平方）)的积分怎么求啊？上下限是从0到二分之派
拜托啦 (cosx)^2

Answer 1

∫ (0,π/2) ln (b²cos²x) dx
=∫ (0,π/2) [ln(b²)+ln(cos²x)]dx
=2xlnb|(0,π/2)+2∫(0,π/2)ln(cosx)dx
考虑第二项
∫(0,π/2)ln(cosx)dx
=∫(π/2,0)ln[cos(π/2-t)d(π/2-t)
=∫(0,π/2)ln(sint)dt
=∫(0,π/2)ln(sinx)dx

∴∫(0,π/2)ln(sinx)dx
=∫(0,π/2)ln(cosx)dx
=∫(0,π/4)ln(cosx)dx+∫(π/4,π/2)ln(cosx)dx
=∫(0,π/4)ln(cosx)dx+∫(π/4,0)ln[cos(π/2-t)]d(π/2-t)
=∫(0,π/4)ln(cosx)dx+∫(0,π/4)ln(sint)dt
=∫(0,π/4)ln(cosx)dx+∫(0,π/4)ln(sinx)dx
=∫(0,π/4) [ln(cosx)+ln(sinx)] dx
=∫(0,π/4) ln[(sin2x)/2] dx
=∫(0,π/4) [ln(sin2x)-ln2] dx
=(1/2)∫(0,π/4) ln(sin2x)d(2x)-(π/4)ln2
=(1/2)∫(0,π/2) ln(sint)dt-(π/4)ln2
=(1/2)∫(0,π/2) ln(sinx)dx-(π/4)ln2
∴∫(0,π/2)ln(sinx)dx
=∫(0,π/2)ln(cosx)dx
=-(π/2)ln2

于是∫ (0,π/2) ln (b²cos²x) dx
=2xlnb|(0,π/2)+2∫(0,π/2)ln(cosx)dx
=πlnb-πln2
=πln(b/2)

Answer 2

是(cosx)^2还是cos(x^2)

追问

(cosx)^2

Answer 3

微积分都忘没了，

追问

擦