∫(sinx)^4dx+∫(sinx)^6dx

1个回答

  • 记A=∫(0到π) x(sinx)^6dx,换元x=π-t,则A=∫(0到π) π(sint)^6dt-∫(0到π) t(sint)^6dt,所以A=π/2×∫(0到π) (sinx)^6dx.

    (sinx)^6以π为周期,且是偶函数,所以∫(0到π) (sinx)^6dx=∫(-π/2到π/2) (sinx)^6dx=2∫(0到π/2) (sinx)^6dx,套用定积分公式,∫(0到π) (sinx)^6dx=2×5/6×3/4×1/2×π/2

    所以,原积分A=π/2×2×5/6×3/4×1/2×π/2=5π^2/32