Question

How do you find the integral of `cos(mx)*cos(nx)`?

Answer 1

`intcos(mx)cos(nx)dx=sin(x(m-n))/(2(m-n))+sin(x(m+n))/(2(m+n))+C`

Explanation:

Your question is:

`intcos(mx)cos(nx)dx`

To simplify this, use the cosine product-to-sum formula, namely:

`cos(A)cos(B)=1/2[cos(A-B)+cos(A+B)]`

Applying this to the cosine functions in the integral, we see that it becomes

`=int1/2[cos(mx-nx)+cos(mx+nx)]dx`

We can split up the integral through addition and do a little internal factoring:

`=1/2intcos(x(m-n))dx+1/2intcos(x(m+n))dx`

Now, we should perform substitution. Focusing for now on just the first integral, we should let `u=x(m-n)` which implies that `du=(m-n)dx`.

In the first integral, multiply the interior by `(m-n)` and balance this by multiplying the outside by `1/(m-n)`.

`1/(2(m-n))intcos(x(m-n))*(m-n)dx`

Using our `u` and `du` values from earlier, this becomes

`=1/(2(m-n))intcos(u)du`

Since `intcos(u)du=sin(u)+C`, this equals

`=1/(2(m-n))sin(u)+C`

`=sin(x(m-n))/(2(m-n))+C`

The second integral can be integrated with the exact same method, except we would set `u=x(m+n)`. It leaves us with the integral:

`1/2intcos(x(m+n))dx=sin(x(m+n))/(2(m+n))+C`

Combing the two integrals, with their respective constants of integration absorbed into one, the final antiderivative is:

`intcos(mx)cos(nx)dx=sin(x(m-n))/(2(m-n))+sin(x(m+n))/(2(m+n))+C`

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Application:

Find

`intcos(13x)cos(8x)dx`

Working through the process very fast, we see

`=int1/2[cos(5x)+cos(21x)]dx`

`=1/2intcos(5x)dx+1/2intcos(21x)dx`

`=1/10intcos(5x)*5dx+1/42intcos(21x)*21dx`

`=sin(5x)/10+sin(21x)/42+C`

We can check this answer using the "formula" we just created:

`intcos(mx)cos(nx)dx=sin(x(m-n))/(2(m-n))+sin(x(m+n))/(2(m+n))+C`

We have `m=13` and `n=8`, so the answer should be:

`sin(x(13-8))/(2(13-8))+sin(x(13+8))/(2(13+8))+C`

`=sin(5x)/10+sin(21x)/42+C`

Confirmed! This matches what we got when we integrated without using our formula.