被积函数可以拆成两部分之和
①xsin^4 x 它是奇函数,∫[-π/2,π/2]xsin^4 xdx=0
②cos²xsin^4 x,它是偶函数,
所以原定积分只需求②的定积分
∫[-π/2,π/2]cos²xsin^4 xdx
=2*∫[0,π/2]cos²xsin^4 xdx
=2*∫[0,π/2] (sin^4 x-sin^6 x)dx
根据公式
∫[0,π/2]sin^n xdx
=∫[0,π/2]cos^n xdx
=
n为偶数时,(n-1)/n*(n-3)/(n-2)*…*1/2*π/2
n为奇数时,n/(n+1)*(n-2)/(n-1)*…*2/3
∫[0,π/2]sin^4 xdx=3/4*1/2*π/2=3π/16
∫[0,π/2]sin^6 xdx
=5/6*3/4*1/2*π/2=15π/96
所以原式=2×3π/96
=π/16
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