We will keep in mind the formula for integration by parts, which is:
#int u dv = uv - int v du#
To find this integral successfully we will let #u = x#, and #dv = cos 5x dx#. Therefore, #du = dx# and #v = 1/5 sin 5x#. (#v# can be found using a quick #u#-substitution)
The reason I chose #x# for the value of #u# is because I know that later on I will end up integrating #v# multiplied by #u#'s derivative. Since the derivative of #u# is just #1#, and since integrating a trig function by itself doesn't make it any more complex, we've effectively removed the #x# from the integrand and only have to worry about the sine now.
So, plugging into the IBP's formula, we get:
#int xcos5x dx = (x sin5x) / 5 - int 1/5 sin 5x dx#
Pulling the #1/5# out of the integrand gives us:
#int xcos5x dx = (x sin5x) / 5 - 1/5 int sin 5x dx#
Integrating the sine will only take a #u#-substitution. Since we've already used #u# for the IBP's formula I'll use the letter #q# instead:
#q = 5x#
#dq = 5 dx#
To get a #5 dx# inside the integrand I'll multiply the integral by another #1/5#:
#int xcos5x dx = (x sin5x) / 5 - 1/25 int 5sin 5x dx#
And, replacing everything in terms of #q#:
#int xcos5x dx = (x sin5x) / 5 - 1/25 int sinq*dq#
We know that the integral of #sin# is #-cos#, so we can finish this integral off easily. Remember the constant of integration:
#int xcos5x dx = (x sin5x) / 5 + 1/25 cos q + C#
Now we will simply substitute back #q#:
#int xcos5x dx = (x sin5x) / 5 + (cos 5x) /25 + C#
And there is our integral.
