哪用这么麻烦?
∫ sinxsin3x dx
= (1/2)∫ [cos(x - 3x) - cos(x + 3x)] dx
= (1/2)∫ cos2x dx - (1/2)∫ cos4x dx
= (1/2)(1/2)sin2x - (1/2)(1/4)sin4x + C
= (1/4)sin2x - (1/8)sin4x + C
公式：sinAsinB = (1/2)[cos(A - B) - cos(A + B)]
哪用这么麻烦?
∫ sinxsin3x dx
= (1/2)∫ [cos(x - 3x) - cos(x + 3x)] dx
= (1/2)∫ cos2x dx - (1/2)∫ cos4x dx
= (1/2)(1/2)sin2x - (1/2)(1/4)sin4x + C
= (1/4)sin2x - (1/8)sin4x + C
公式：sinAsinB = (1/2)[cos(A - B) - cos(A + B)]