不定积分∫sinxcos^2xdx分部积分法

Question

如题所述

Answer 1

∫sinxcos^2xdx=∫sinxcosxdsinx=1/2∫cosxdsin^2x
=1/2cosxsin^2x+1/2∫sin^3x=1/2cosxsin^2x+1/2∫sinx(1-cos^2x)dx
=1/2cosxsin^2x-1/2cosx-1/2∫sinxcos^2xdx
故：∫sinxcos^2xdx=2/3[1/2cosxsin^2x-1/2cosx]+C