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

1 Answer
Mar 26, 2016

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.