江画工太猩院 例5设y=sin(2x+1,求 解∵ y=sin u, u=2x+1. =c0dt=c0s(2x+1)d(2x+1) =C0(2x+1)·2x=20(2x+1)tx 例6设y=e" sin bx,.求l 解dy=e-· cos bxd(bx)+ sin bx e d(-ax) =ea, cos bx·bd+ sin bx·e -lX (-a) e (b cos bx -a sin bx dx.江西理工大学理学院 例6 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5 解 设 y = sin(2x + 1),求dy. Q y = sin u, u = 2x + 1. ∴dy = cos udu = cos(2x + 1)d(2x + 1) = cos(2x + 1)⋅ 2dx = 2cos(2x + 1)dx