Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 46

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 sin(5theta)sin(4theta) d theta has an integrand in the form of a trigonometric function. To evaluate this, we apply the identity:
sin(A)sin(B) =[-cos(A+B) +cos(A-B)]/2
The integral becomes:
intsin(5theta)sin(4theta)d theta= int[-cos(5theta+4theta) + cos(5theta -4theta)]/2 d theta
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int[-cos(5theta+4theta) + cos(5theta -4theta)]/2 d theta= 1/2int[-cos(5theta+4theta) +cos(5theta -4theta)] d theta
Apply the basic integration property:int (u+v) dx = int (u) dx + int (v) dx .
1/2 *[int-cos(5theta+4theta)d theta+cos(5theta -4theta)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+4theta) d theta, we let u =5theta +4theta =9theta then du= 9 d theta or (du)/9 =d theta .
int -cos(5theta+4theta)d theta=int -cos(9theta)d theta
=int -cos(u) *(du)/9
= -1/9 int cos(u)du
= -1/9 sin(u) +C
Plug-in u =9theta on -1/9 sin(u) +C , we get:
int-cos(5theta+4theta)d theta= -1/9 sin(9theta) +C
For the integral: intcos(5theta -4theta)d theta , we let u =5theta -4theta =theta then du= d theta .
intcos(5theta -4theta)d theta = intcos(theta) d theta
=intcos(u) *(du)
=sin(u) +C
Plug-in u =theta on 1/2 sin(u) +C , we get:
intcos(5theta -4theta)d theta = sin(theta) +C
Combining the results, we get the indefinite integral as:
intsin(5theta)sin(4theta)d theta = 1/2*[ -1/9 sin(9theta) + sin(theta)] +C
or - 1/18 sin(4theta) +1/2 sin(theta) +C