Question

sin^2xcos^3x的不定积分，求过程

Answer 1

∫ sin²x cos³x dx
=∫ sin²x cos²x d[sinx]
=∫ sin²x (1 - sin²x) d[sinx]
=∫ (sin²x - [sinx]^4) d[sinx]
=(1 / 3)sin³x - (1 / 5)[sinx]^5 + C

Answer 2

∫sin^2x(1-sin^2x)d(sinx)=∫sin^2xd(sinx)-∫sin^4xd(sinx)=(1/3)∫d(sin^3x)-(1/5)∫d(sin^5x)=(1/3)sin^3x-(1/5)sin^2x+C

Answer 3

∫sin^2xcos^3xdx = ∫sin^2x (1-sin^2x)dsinx= ∫sin^2x-sin^4x dx = (1/3) sin^3x-(1/5)sin^5x+C
实际非常简单

追问

那么sin2xcos3x 的不定积分呢