Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 45

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