Question

高等数学，求详细过程

Answer 1

被积函数是偶函数
原式=2*∫[0，1](x²+1)√(1-x²)dx
=2*∫[0,1]x²√(1-x²)dx+2∫[0,1]√(1-x²)dx
∫[0,1]x²√(1-x²)dx
=∫[0,π/2]sin²tcos²tdt
=1/4*∫[0,π/2]sin²2tdt
=1/4*∫[0,π/2](1-cos4t)/2dt
=(1/8*t-1/32*sin4t)|[0,π/2]
=π/16
∫[0,1]√(1-x²)dx
=∫[0,π/2]cos²tdt
=∫[0,π/2](1+cos2t)/2dt
=(1/2*t+1/4*sin2t)|[0,π/2]
=π/4
所以原式=2×(π/16+π/4)=5π/8

Answer 2

令 x = sinu， 积分函数是偶函数，则
I = 2 ∫<0, π/2> [1+(sinu)^2](cosu)^2du
= 2 ∫<0, π/2> [(cosu)^2+(1/4)(sin2u)^2]du
= ∫<0, π/2> [1+cos2u+(1/4)(1-cos4u)]du
= [(5/4)u + (1/2)sin2u - (1/16)sin4u]<0, π/2> = 5π/8

Answer 3

令x=sinu 上下限为pi/2,-pi/2
代入得=定积分（(sinx）^2+1)(cosu)^2du

＝定积分((sin2u)^2/4+(cosu)^2)du
=定积分((1-cos4u)/8+(1+cos2u)/2)du
=(5u/8+sin2u/4-sin4u/32)
=5pi/8+5/16

Answer 4

追问

不对，答案等于派/2

追答

我没看出我哪错了

题没抄错吧

追问

可能是答案错了吧

追答

好吧