int tan^6(3x) dx Find the indefinite integral

To evaluate the integral int tan^6(3x) dx , we apply u-substitution by letting:
u =3x then  du = 3 dx or  (du)/3 = dx .
Plug-in the values, we get:
int tan^6(3x) dx =int tan^6(u) * (du)/3
 Apply the basic properties of integration: int c*f(x) dx= c int f(x) dx .
int tan^6(u) * (du)/3= 1/3int tan^6(u) du
Apply integration formula for tangent function: int tan^n(x)dx = (tan^(n-1)(x))/(n-1)- int tan^(n-2)(x)dx .
1/3int tan^6(u) du =1/3 *[(tan^(6-1)(u))/(6-1)- int tan^(6-2)(u)du]
                             =1/3*[(tan^(5)(u))/(5)- int tan^(4)(u)du]
Apply another set integration formula for tangent function on int tan^(4)(u)du .
int tan^(4)(u)du =(tan^(4-1)(u))/(4-1)- int tan^(4-2)(u)du
                          =(tan^(3)(u))/(3)- int tan^(2)(u)du
For the integral of int tan^(2)(u)du , we may apply integration formula: int tan^2(x) dx = tan(x)-x+C .
int tan^(2)(u)du =tan(u)-u +C
 
Applying int tan^(2)(u)du =tan(u)-u +C , we get:
int tan^(4)(u)du =(tan^(3)(u))/(3)- int tan^(2)(u)du
                         =(tan^(3)(u))/(3)- [tan(u)-u] +C
                          =(tan^(3)(u))/(3)- tan(u)+u +C
Applying  int tan^(4)(u)du=(tan^(3)(u))/(3)- tan(u)+u +C . we get:
1/3int tan^6(u) du=1/3*[(tan^(5)(u))/(5)- int tan^(4)(u)du]
                              =1/3*[(tan^(5)(u))/(5)- [(tan^(3)(u))/(3)- tan(u)+u]] +C
                             =1/3*[(tan^(5)(u))/(5)- (tan^(3)(u))/(3)+ tan(u)-u] +C
                             = (tan^(5)(u))/15- (tan^(3)(u))/9+ tan(u)/3-u/3 +C
Plug-in u = 3x on (tan^(5)(u))/15- (tan^(3)(u))/9+ tan(u)/3-u/3 +C ,we get the indefinite integral as:
int tan^6(3x) dx=(tan^(5)(3x))/15-( tan^(3)(3x))/9+ tan(3x)/3-(3x)/3 +C
                          =(tan^(5)(3x))/15-( tan^(3)(3x))/9+ tan(3x)/3-x +C