解：∫sin3XsinXdx =∫-（1/2)[cos(3x+x)-cos(3x-x)]dx =-(1/2)∫（cos4x-cos2x）dx =（-1/2）[(1/4)sin4x-(1/2)sin2x]+C =-(1/8)sin12x+(1/4)sin2x+C不懂再问懂请采纳
