Question

∫cos3xcos2xdx怎样求它的定积分啊……

要具体步骤，麻烦高手详细讲解一下

Answer 1

不用他们那么麻烦!
积化和差:cosαcosβ=[cos（α+β）+cos（α-β）]/2
cos3xcos2x=[cos5x+cosx]/2
原式=1/10sin5x+1/2sinx+C

Answer 2

cos3xcos2x=1/2[cos5x+cosx]
所以,
∫cos3xcos2xdx=1/2*∫(cos5x+cosx)dx
=1/10*sin5x+1/2*sinx+C
C表示任意常数

Answer 3

∫cos3xcos2xdx
=(1/2)∫cos3xdsin2x
= (1/2)cos3xsin2x + (3/2)∫sin3xsin2xdx
=(1/2)cos3xsin2x - (3/4)∫sin3xdcos2x
=(1/2)cos3xsin2x - (3/4)sin3xcos2x +(9/4)∫cos3xcos2x dx
(-5/4)∫cos3xcos2xdx=(1/2)cos3xsin2x - (3/4)sin3xcos2x
∫cos3xcos2xdx=(-4/5)[(1/2)cos3xsin2x - (3/4)sin3xcos2x]+C

Answer 4

∫cos3xcos2xdx

=1/2∫(cos5x-cosx)dx
=1/10sin5x-1/2sinx+C