Question

求不定积分

Answer 1

I = ∫(-π/2->π/2) (e^x)sin⁴x/(1+e^x) dx
令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)
代值进去~
=3π/16

追问

有乱码看不清

追答

……晕。用电脑看没乱码……

Answer 2

这么复杂，让我问问告你。