First let #u = 3x, du = 3dx# to simplify the integral. Then we have:
#1/3##intcos^5(u)du#
Since the power of cos(u) is odd, we can use the procedure of saving one #cos(u)# and using the identity #cos(u)^2=(1-sin^2(u))#
#=># #1/3##int(1-sin^2(u))^2cos(u)du#
Remember we made it #(1-sin^2(u))^2# all squared because our original function is #cos(u)^4#
After this we need to use another substitution. For this time: let #z =
sin(u)#, #dz = cos(u)du#
#=># #1/3##int(1-z^2)^2dz#
Now it's a simple polynomial integral we can evaluate it straightforwardly
#=# #1/3##int(1-2z^2+z^4)dz#
#=1/3(z-2z^3/3+z^5/5)+C#
We started with #x# we substitute back to #x#
#=1/3(sin(u)-2(sin(u))^3/3+(sin(u))^5/5)+C#
#=1/3(sin(3x)-2(sin(3x))^3/3+(sin(3x))^5/5)+C#
