Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 42

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(5theta)cos(3theta) d theta has an integrand in a form of a trigonometric function. To evaluate this, we apply the identity:
cos(X)cos(Y) =[cos(X+Y) +cos(X-Y)]/2
The integral becomes:
int cos(5theta)cos(3theta) d theta = int[cos(5theta+3theta) + cos(5theta -3theta)]/2 d theta
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int[cos(5theta+3theta) + cos(5theta -3theta)]/2d theta = 1/2int[cos(5theta+3theta) + cos(5theta -3theta)] d theta
Apply the basic integration property: int (u+v) dx = int (u) dx + int (v) dx .
1/2 *[int cos(5theta +3theta)d theta+cos(5theta -3theta)d theta]
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(5theta +3theta)d theta , we let u =5theta +3theta =8theta then du= 8 d theta or (du)/8 =d theta .
int cos(5theta +3theta)d theta=int cos(8theta)d theta
=intcos(u) *(du)/8
= 1/8 int cos(u)du
= 1/8 sin(u) +C
Plug-in u =8theta on 1/8 sin(u) +C , we get:
int cos(5theta +3theta)d theta=1/8 sin(8theta) +C
For the integral: intcos(5theta -3theta)d theta , we let u =5theta -3theta =2theta then du= 2 d theta or (du)/2 =d theta .
int cos(5theta -3theta)d theta = intcos(2theta) d theta
=intcos(u) *(du)/2
= 1/2 int cos(u)du
= 1/2 sin(u) +C
Plug-in u =2 theta on 1/2 sin(u) +C , we get:
intcos(5theta -3theta)d theta =1/2 sin(2theta) +C
Combing the results, we get the indefinite integral as:
int cos(5theta)cos(3theta)d theta = 1/2*[1/8 sin(8theta) +1/2 sin(2theta)] +C
or 1/16 sin(8theta) +1/4 sin(2theta) +C