例3设y=sin(2x+1),求小 解 ∵y≡sin,l=2x+1. cos udu =cos(2x +1d(2x+1) =c0s(2x+1)·2x=2cos(2x+1)dx 例4设y= e sin bx,求 解d cos bxd (br) +sin bx.e d(ax) e cos bx bdx+ sin bx..cadx e (b cos bx-a sin bx)dx. 1313 例 4 解 y e sin bx, dy. 设 = −ax 求 dy e cos bxd(bx) sin bx e d( ax) ax ax =  +  − − − e bx bdx bx e a dx ax ax =  cos  + sin  (− ) − − e (bcos bx a sin bx)dx. ax = − − 例 3 解 设 y = sin( 2 x + 1), 求dy.  y = sin u, u = 2x + 1. dy = cos udu = cos(2x + 1)d(2x + 1) = cos(2x + 1) 2dx = 2cos(2x + 1)dx