Question

sin^4x（e^x/1+e^x）的定积分，上限是π/2，下限是-π/2

Answer 1

换元积分法，把x换成-t，得到一个新积分，它与原积分是等值的。新积分与原积分相加，化简，可得一个关于sin^4x的定积分，积出来后除以2就是原积分

Answer 2

I = ∫(-π/2->π/2) (e^x)sin⁴x/(1+e^x) dx
Let x = -y，dx = -dy
=> ∫(π/2->-π/2) (e^-y)sin⁴(-y) / (1+e^-y) (-dy)
= ∫(-π/2->π/2) sin⁴y/(1+e^y) dy
= ∫(-π/2->π/2) sin⁴x/(1+e^x) dx = J

I = (1/2)(I+J)
= (1/2)[∫(-π/2->π/2) (e^x)sin⁴x/(1+e^x) dx + ∫(-π/2->π/2) sin⁴x/(1+e^x) dx]
= (1/2)∫(-π/2->π/2) (1+e^x)sin⁴x/(1+e^x) dx
= (1/2)∫(-π/2->π/2) sin⁴x dx
= (2)(1/2)∫(0->π/2) [(1-cos2x)/2)]² dx
= (1/4)∫(0->π/2) (1-2cos2x+cos²2x) dx
= (1/4)∫(0->π/2) (1-2cos2x) dx + (1/8)∫(0->π/2) (1+cos4x) dx
= (1/4)(x - sin2x) + (1/8)(x + 1/4 * sin4x)
= (1/4)(π/2 - sinπ) + (1/8)(π/2 + 1/4 * sin2π)
= π/8 + π/16
= 3π/16

Answer 3

个人认为，你也可以参考下分步积分法