求下列积分∫(0到1)x(1-x^4)^3/2dx

Question

求下列积分∫(0到1)x(1-x^4)^3/2dx

Answer 1

∫[0→1] x(1-x^4)^(3/2) dx
=(1/2)∫[0→1] (1-x^4)^(3/2) d(x²)
令x²=y
=(1/2)∫[0→1] (1-y²)^(3/2) dy
令y=sinu,则(1-y²)^(3/2)=cos³u,dy=cosudu,u：0→π/2
=(1/2)∫[0→π/2] (cosu)^4 du
=(1/8)∫[0→π/2] (1+cos2u)² du
=(1/8)∫[0→π/2] (1+2cos2u+cos²2u) du
=(1/8)∫[0→π/2] [1+2cos2u+(1/2)(1+cos4u)] du
=(1/8)[3u/2 + sin2u + (1/8)sin4u] |[0→π/2]
=3π/32

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