Recall that indefinite integral follows `int f(x) dx = F(x) +C ` where:

`f(x) ` as the integrand function

`F(x)` as the antiderivative of f(x)

`C` as the constant of integration..

For the given integral problem: `int cos^2(3x) dx` , we can evaluate this by using a trigonometric identity. Recall...

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Recall that indefinite integral follows `int f(x) dx = F(x) +C ` where:

`f(x) ` as the integrand function

`F(x)` as the antiderivative of f(x)

`C` as the constant of integration..

For the given integral problem: `int cos^2(3x) dx` , we can evaluate this by using a trigonometric identity. Recall that:

`cos^2(theta) = (1 + cos(2theta))/2` .

Applying the trigonometric identity, we get:

`int cos^2(3x) dx = int (1 + cos(2* 3x))/2 dx`

`= int ( 1 + cos(6x))/2dx`

`=int ( 1/2 + cos(6x)/2)dx`

Apply the basic integration property: : `int (u+v) dx = int (u) dx + int (v) dx` .

`int ( 1/2) + cos(6x)/2)dx =int ( 1/2) dx + int cos(6x)/2dx`

For the first integral: `int (1/2) dx` , we may apply basic integration property: `int c dx = cx` .

`int (1/2) dx = 1/2x or x/2`

For the second integral: `int cos(6x)/2dx` , we may apply basic integration property: `int c f(x) dx = c int f(x) dx` .

`1/2 int cos(6x) dx` .

Apply u-substitution by letting `u = 6x` then `du = 6 dx` or `(du)/6 = dx` .

`1/2 int cos(6x) dx = 1/2 int cos(u) * (du)/6`

`= 1/2*1/6 int cos(u) du`

`= 1/12 sin(u)`

Plug-in `u = 6x` on `1/12sin(u)` , we get:

`1/2 int cos(6x) dx = 1/12 sin(6x) ` or `sin(6x)/12`

Combining the results, we get the indefinite integral as:

`int cos^2(3x) dx = x/2 + sin(6x)/12+C`