Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 10

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
To evaluate the given problem int sin^4(6theta) d theta , we may apply u-substitution by letting: u = 6theta then du = 6 d theta or (du)/6 = d theta .
The integral becomes:
int sin^4(6theta) d theta=int sin^4(u) * (du)/6
Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int sin^4(u) * (du)/6=1/6int sin^4(u)du .
Apply the integration formula for sine function: int sin^n(x) dx = -(cos(x)sin^(n-1)(x))/n+(n-1)/n int sin^(n-2)(x)dx .
1/6int sin^4(u)du=1/6[-(cos(u)sin^(4-1)(u))/4+(4-1)/4 int sin^(4-2)(u)du] .
=1/6[-(cos(u)sin^(3)(u))/4+3/4 int sin^(2)(u)du]
For the integral int sin^(2)(u)du , we may apply trigonometric identity: sin^2(x)= 1-cos(2x)/2 or 1/2 - cos(2x)/2.
We get:
int sin^(2)(u)du = int ( 1/2 - cos(2u)/2) du .
Apply the basic integration property:int (u-v) dx = int (u) dx - int (v) dx .
int ( 1/2 - cos(2u)/2) du=int ( 1/2) du - int cos(2u)/2 du
= 1/2u - 1/4sin(2u)+C
or u/2 - sin(2u)/4+C
Note: From the table of integrals, we have int cos(theta) d theta = sin(theta)+C.
Let: v = 2u then dv = 2du or (dv)/2= du
thenint cos(2x)/2 du =int cos(v)/2 * (dv)/2
= 1/4 sin(v)
= 1/4 sin(2u)
Applying int sin^(2)(u)du=u/2 - sin(2u)/4+C , we get:
1/6int sin^4(u)du=1/6[-(cos(u)sin^(3)(u))/4+3/4 int sin^(2)(u)du]
=1/6[-(cos(u)sin^(3)(u))/4+3/4 [u/2 - sin(2u)/4]]+C
=1/6[-(cos(u)sin^(3)(u))/4+(3u)/8 - (3sin(2u))/16]+C
=(-cos(u)sin^(3)(u))/24+(3u)/48 - (3sin(2u))/96+C
Plug-in u =6theta on (-cos(u)sin^(3)(u))/24+(3u)/48 - (3sin(2u))/96+C to find the indefinite integral as:
int sin^4(6theta) d theta =(cos(6theta)sin^(3)(6theta))/24+(3*6theta)/48 - (3sin(2*6theta))/96+C
=(cos(6theta)sin^(3)(6theta))/24+(18theta)/48 - (3sin(12theta))/96+C
=(cos(6theta)sin^(3)(6theta))/24+(3theta)/8 - (sin(12theta))/32+C