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

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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`


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