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

2 Answers
Apr 23, 2018

1/14sin7t+1/2sint+C

Explanation:

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

Apr 23, 2018

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