How do you integrate #int cos(3t)cos(4t)dt#?

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Answer 1 · 2018-04-23T18:10:32

#1/14sin7t+1/2sint+C#

Recall the cosine product identity

#cosacosb=1/2[cos(a+b)+cos(a-b)]#

Using it, we can say

#cos3tcos4t=1/2[cos(3t+4t)+cos(3t-4t)]#

#=1/2(cos7t+cos(-t))#

Cosine is an even function, so #cos(-t)=cost#

#=1/2(cos7t+cost)#

Thus, we get

#1/2int(cos7t+cost)dt=1/2intcos7tdt+1/2intcostdt#

#=1/14sin7t+1/2sint+C#

Answer 2 · 2018-04-23T18:17:57

Integrate: #int cos(3t)cos(4t)dt#

Use the identity #cos(A)cos(B) = 1/2(cos(A-B)+cos(A+B))# where #A = 4t# and #B = 3t#:

#int cos(3t)cos(4t)dt = 1/2intcos(t) +cos(7t)dt#

Separate into two integrals:

#int cos(3t)cos(4t)dt = 1/2intcos(t)dt +1/2intcos(7t)dt#

The two integrals are trivial:

#int cos(3t)cos(4t)dt = 1/2sin(t) +1/14sin(7t)+ C#