Question

考试中，在线等

考试中，在线等急急急

Answer 1

∫sin2xcos3xdx
=½∫[sin(3x+2x)-sin(3x-2x)]dx
=½∫(sin5x-sinx)dx
=(-1/10)cos5x +½cosx +C

追答

sin(A)cos(B) = (1/2)[sin(A+B) + sin(A-B)]
∴sin(2x)cos(3x) = (1/2)[sin(2x+3x) + sin(2x-3x)]
= (1/2)[sin5x + sin(-x)]
= (1/2)(sin5x - sinx)
∴∫ sin(2x)cos(3x) dx
= (1/2)∫ sin(5x) dx - (1/2)∫ sinx dx
= (1/2)(1/5)∫ sin(5x) d(5x) - (1/2)∫ sinx dx
= (1/10)(-cos(5x)] + (1/2)cosx + C
= (1/10)[5cosx - cos(5x)] + C

二个，，不知那个对的

Answer 2

∫（sin2x+cos3x)dx
=∫sin2xdx+∫cos3xdx
=1/2∫sin2xd(2x)+1/3∫cos3xd(3x)
=-1/2*cos2x+1/3*sin3x+C
(C是常数)