Question

帮忙求一下这两个积分怎么计算？

Answer 1

(1)
∫sin(mx)·sin(nx) dx
=½∫[cos(m-n)x-cos(m+n)x]dx
= -1/[2(m+n)]∫cos(m+n)x d[(m+n)x]+1/[2(m-n)]∫cos(m-n)x d[(m-n)x]
=sin[(m+n)x]/[2(m+n)] -sin[(m-n)x]/[2(m-n)] +C
(2)
∫cosxcos²(3x) dx
=∫cosx[1+cos(6x)]/2 dx
=½∫cosx dx +½∫cosxcos(6x)dx
=½∫cosx dx +¼∫[cos(7x)+cos(5x)]dx
=½∫cosx dx +(1/28)∫[cos(7x)dx+ (1/20)∫cos(5x)]d(5x)
=-½sinx -(1/28)sin(7x)-(1/20)sin(5x) +C

两题都需要通过三角函数恒等变形进行化简。

Answer 2

(1)
∫sin(mx).sinnx dx
=(1/2)∫[ cos(m-n)x - cos(m+n)x ] dx
=(1/2)[ (1/(m-n)). sin(m-n)x -(1/(m+n)).sin(m+n)x ] +C

(2)

∫cosx.(cos3x)^2 dx
=(1/2)∫cosx.(1+cos6x) dx
=(1/2)sinx +(1/2)∫cosx.cos6x dx
=(1/2)sinx +(1/4)∫(cos5x+cos7x) dx
=(1/2)sinx +(1/4)[ (1/5)sin5x+(1/7)sin7x ] + C