If you just want an answer, there is always a table of trig integrals.
If you want to see details of the integration, let cos3x = cosx(1 - sin2x)
Then the 2 integrals will be of cosx and -cosxsin2x
the first integral is easy (using the constant 19) to be 19sinx
for the second integral, you let u = sin2x and du = 2sinxcosx dx giving the integral of u2 du
the solution for this integral (using the 19 constant) will be -19sin3x
Put them together for the total answer adding a general constant to complete the answer of
19sinx - (19/3)sin3x + constant
