Question

不明白sin^4θ-sin^6θ怎么计算得下面的结果 麻烦来个详细过程谢谢

Answer 1

直接用华里士公式即可

Answer 2

I(2n)
=∫(0->π/2) (sinx)^(2n) dx
=-∫(0->π/2) (sinx)^(2n-1) dcosx
=-[cosx.(sinx)^(2n-1)]|(0->π/2) +(2n-1)∫(0->π/2) (sinx)^(2n-2) (cosx)^2 dx
=0 +(2n-1)∫(0->π/2) (sinx)^(2n-2) [1-(sinx)^2] dx
2nI(2n) =(2n-1)I(2n-2)
I(2n) =[(2n-1)/(2n)]I(2n-2)
=>
∫(0->π/2) (sinx)^6 dx
=I6
=(5/6)I4
=(5/6)(3/4)I2
=(5/6)(3/4)(1/2)I0
=(5/6)(3/4)(1/2)(π/2)
∫(0->π/2) (sinx)^4 dx
=(3/4)I2
=(3/4)(1/2)I0
=(3/4)(1/2)(π/2)