∫（sin2xcos3x）dx用换元法求积分怎么求啊!

∫（sin2xcos3x）dx用换元法求积分怎么求啊!
∫（sin2xcos3x）dx用换元法求积分怎么求啊!

∫（sin2xcos3x）dx用换元法求积分怎么求啊!
∫（sin2xcos3x）dx
=ʃ1/2(sin5x-sinx)dx
=1/2[ʃsin5xdx-ʃsinxdx]
=1/2[1/5*ʃsin5xd5x+cosx]
=1/10(-cos5x)+1/2cosx+C
=-1/10*cos5x+1/2cosx+C

sin2x * cos3x = (1/2) * [sin(2x+3x) + sin(2x-3x)]
= (1/2) * [sin5x + sin(-x)]
= (1/2) * (sin5x - sinx)
∫ sin2x * cos3x dx
= (1/2)∫ (sin5x - sinx) dx
= (1/2) * [(-1/5)cos5x - (-cosx)] + c
= (1/10) * (5cosx - cos5x) + c

I=∫（sin2xcos3x）dx
= (-1/2)∫ cos3x dcos2x
= -cos3xcos2x/2 + (1/2) ∫ cos2x sin3x dx
= -cos3xcos2x/2 + (1/4) ∫ sin3x dsin2x
= -cos3xcos2x/2 + (1/4) sin2x sin3x +(1/12)∫ sin2x cos3x dx
= = -cos3xcos2x/2 + (1/4) sin2x sin3x +(1/12)I
I = (12/11)(-cos3xcos2x/2 + (1/4) sin2x sin3x) + C