sin(x/3)sin(nx)dx的不定积分

Question

sin(x/3)sin(nx)dx的不定积分

Answer 1

积化和差公式：
sin(x/3)sin(nx) = (1/2)[cos(x/3 - nx) - cos(x/3 + nx)]
∫ sin(x/3)sin(nx) dx
= (1/2)∫ cos(x/3 - nx) dx - (1/2)∫ cos(x/3 + nx) dx
= (1/2)[1/(1/3 - n)]∫ cos[(1/3 - n)x] d[(1/3 + n)x] - (1/2)[1/(1/3 + n)]∫ cos[(1/3 + n)x] d[(1/3 + n)x]
= 1/[2(1/3 - n)]sin(x/3 - nx) - 1/[2(1/3 + n)]sin(x/3 + nx) + C
= [3/(2 - 3n)]sin(x/3 - nx) - [3/(2 + 3n)]sin(x/3 + nx) + C