帮我解一道数学题

a)

I(n)
=∫(sinx)^ndx
=∫(sinx)^(n-2)(1-(cosx)^2)dx
=I(n-2)-∫(sinx)^(n-2)(cosx)^2dx
=I(n-2)-∫(sinx)^(n-2)cosxd(sinx)
=I(n-2)+∫(sinx)^(n-1)d(cosx)/(n-1) -(sinx)^(n-1)cosx/(n-1)|(上下限轮野，下同）
=I(n-2)-∫(sinx)^ndx/(n-1) -(sinx)^(n-1)cosx/(n-1)|
=I(n-2)-I(n)/(n-1) -(sinx)^(n-1)cosx/(n-1)|

I(n) = I(n-2)-I(n)/(n-1) -(sinx)^(n-1)cosx/(n-1)|

I(n) = ((n-1)/n)I(n-2) - (sinx)^(n-1)cosx/n|(上下限）

b)

I(n-2) = (n/(n-1))I(n) + (sinx)^(n-1)cosx/(n-1)|

I(-2) (π/4----〉π/2，下同）
= 0 - cotx|
= 0 - cotπ/2 + cotπ/4
= 1

∫1/(sin⁴(x))dx (π/4----〉π/2，下同）兄茄
= I(-4)
= (2/腊尘喊3)I(-2) - cosx/3(sinx)^3|
= 2/3 - cos(π/2)/3sin(π/2)^3 + cos(π/4)/3sin(π/4)^3
= 2/3 - 0 + 2/3
= 4/3