Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 9

Recall that indefinite integral follows int f(x) dx = F(x) +C where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration..
For the given integral problem: int cos^2(3x) dx , we can evaluate this by using a trigonometric identity. Recall that:
cos^2(theta) = (1 + cos(2theta))/2 .
Applying the trigonometric identity, we get:
int cos^2(3x) dx = int (1 + cos(2* 3x))/2 dx
= int ( 1 + cos(6x))/2dx
=int ( 1/2 + cos(6x)/2)dx

Apply the basic integration property: : int (u+v) dx = int (u) dx + int (v) dx .
int ( 1/2) + cos(6x)/2)dx =int ( 1/2) dx + int cos(6x)/2dx
For the first integral: int (1/2) dx , we may apply basic integration property: int c dx = cx .
int (1/2) dx = 1/2x or x/2
For the second integral: int cos(6x)/2dx , we may apply basic integration property: int c f(x) dx = c int f(x) dx .
1/2 int cos(6x) dx .
Apply u-substitution by letting u = 6x then du = 6 dx or (du)/6 = dx .
1/2 int cos(6x) dx = 1/2 int cos(u) * (du)/6
= 1/2*1/6 int cos(u) du
= 1/12 sin(u)
Plug-in u = 6x on 1/12sin(u) , we get:
1/2 int cos(6x) dx = 1/12 sin(6x) or sin(6x)/12
Combining the results, we get the indefinite integral as:
int cos^2(3x) dx = x/2 + sin(6x)/12+C