How do you find the integral of #cos(mx)*cos(nx)#?
1 Answer
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
In the first integral, multiply the interior by
#1/(2(m-n))intcos(x(m-n))*(m-n)dx#
Using our
#=1/(2(m-n))intcos(u)du#
Since
#=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
#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
#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.