Question

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

Answer 1

`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`

Answer 2

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`