Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 43

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(2x)cos(4x) dx or intcos(4x)sin(2x) dx has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:
cos(A)sin(B) =[sin(A+B) -sin(A-B)]/2
The integral becomes:
int cos(4x)sin(2x) dx = int[sin(4x+2x) -sin(4x-2x)]/2dx
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int [sin(4x+2x) -sin(4x-2x)]/2dx = 1/2int[sin(4x+2x) -sin(4x-2x)]dx
Apply the basic integration property: int (u+v) dx = int (u) dx + int (v) dx .
1/2 *[int sin(4x+2x)dx+int sin(4x-2x)dx]
Then apply u-substitution to be able to apply integration formula for cosine function: int sin(u) du= -cos(u) +C .
For the integral:int sin(4x+2x)dx , we let u = 4x+2x =6x then du= 6 dx or (du)/6 =dx .
int sin(4x+2x)dx=intsin(6x) dx
=intsin(u) *(du)/6
= 1/6 int sin(u)du
=-1/6cos(u) +C
Plug-in u =6x on -1/6 cos(u) +C , we get:
int sin(4x+2x)dx= -1/6 cos(6x) +C
For the integral: intsin(4x-2x)dx , we let u = 4x-2x =2x then du= 2 dx or (du)/2 =dx .
intsin(4x-2x)dx=intsin(2x) dx
=intsin(u) *(du)/2
= 1/2 int sin(u)du
= -1/2cos(u) +C
Plug-in u =2x on -1/2 cos(u) +C , we get:
intsin(4x-2x)dx= -1/2 cos(2x) +C
Combing the results, we get the indefinite integral as:
intcos(4x)sin(2x) dx= 1/2*[ -1/6 cos(6x) -(-1/2 cos(2x))] +C
or -1/12 cos(6x) +1/4 cos(2x) +C