导数sin2xcos3x的原函数
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I=∫sin2xcos3xdx
=(1/3)∫sin2xdsin3x
=(1/3)sin2xsin3x-(1/3)∫sin3xdsin2x
=(1/3)sin2xsin3x-(2/3)∫sin3x*cos2xdx
=(1/3)sin2xsin3x+(2/9)∫cos2xdcos3x
=(1/3)sin2xsin3x+(2/9)cos2xcos3x-(2/9)∫cos3xdcos2x
=(1/3)sin2xsin3x+(2/9)cos2xcos3x+(4/9)∫cos3xsin2xdx
=(1/3)sin2xsin3x+(2/9)cos2xcos3x+(4/9)I
所以：
I-(4/9)I=(1/3)sin2xsin3x+(2/9)cos2xcos3x
I=(3/5)[sin2xsin3x+(2/3)cos2xcos3x].
=(1/3)∫sin2xdsin3x
=(1/3)sin2xsin3x-(1/3)∫sin3xdsin2x
=(1/3)sin2xsin3x-(2/3)∫sin3x*cos2xdx
=(1/3)sin2xsin3x+(2/9)∫cos2xdcos3x
=(1/3)sin2xsin3x+(2/9)cos2xcos3x-(2/9)∫cos3xdcos2x
=(1/3)sin2xsin3x+(2/9)cos2xcos3x+(4/9)∫cos3xsin2xdx
=(1/3)sin2xsin3x+(2/9)cos2xcos3x+(4/9)I
所以：
I-(4/9)I=(1/3)sin2xsin3x+(2/9)cos2xcos3x
I=(3/5)[sin2xsin3x+(2/3)cos2xcos3x].
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