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

Question

Charles Anctil · Accepted Answer

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

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#

#" "#

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.

Isaiah Allen · Answer

undefined To find the integral of cos(mx) * cos(nx), where m and n are constants:

1. If m is not equal to n, the integral evaluates to 0.
2. If m is equal to n, the integral evaluates to (1/2) * integral of [cos(2mx)] dx.
3. Apply the double angle identity: cos(2mx) = (1/2) * (cos(2mx) + cos(0)), where cos(0) = 1.
4. Integrate cos(2mx) + cos(0) with respect to x.
5. The integral of cos(2mx) is (1/2m) * sin(2mx).
6. The integral of cos(0) is x.
7. Substitute the results back into the expression.
8. Simplify the expression to get the final integral result.