Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 41

Indefinite integrals are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
For the given problem: int cos(6x)cos(2x) dx has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:
cos(A)cos(B) =[cos(A+B) +cos(A-B)]/2
The integral becomes:
int cos(6x)cos(2x) dx = int[cos(6x+2x) +cos(6x-2x)]/2dx

Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int[cos(6x+2x) +cos(6x-2x)]/2dx = 1/2int[cos(6x+2x) +cos(6x-2x)]dx
Apply the basic integration property: int (u+v) dx = int (u) dx + int (v) dx .
1/2 *[intcos(6x+2x) dx+int cos(6x-2x)dx]
Then apply u-substitution to be able to apply integration formula for cosine function: int cos(u) du= sin(u) +C .
For the integral: int cos(6x+2x) dx, we let u = 6x+2x =8x then du= 8 dx or (du)/8 =dx .
intcos(6x+2x) dx=intcos(8x) dx
=intcos(u) *(du)/8
= 1/8 int cos(u)du
= 1/8 sin(u) +C
Plug-in u =8x on 1/8 sin(u) +C , we get:
intcos(6x+2x) dx=1/8 sin(8x) +C
For the integral: intcos(6x-2x) dx , we let u = 6x-2x =4x then du= 4 dx or (du)/4 =dx .
intcos(6x-2x) dx=intcos(4x) dx
=intcos(u) *(du)/4
= 1/4 int cos(u)du
= 1/4 sin(u) +C
Plug-in u =4x on 1/4 sin(u) +C , we get:
intcos(6x-2x) dx=1/4 sin(4x) +C
Combing the results , we get the indefinite integral as:
1/2 *[intcos(6x+2x) dx+int cos(6x-2x)dx] = 1/2*[1/8 sin(8x) +1/4 sin(4x)] +C
or 1/16 sin(8x) +1/8 sin(4x) +C