Before we can begin to find the integral, we must do some rearranging using some trig identities so that we have integrals that we can work with. We will need to remember the following identities as we work through this problem:

`tan^2(x)=sec^2(x)-1`

`tan(x)=(sin(x))/(cos(x))`

So, to begin, we will split up...

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Before we can begin to find the integral, we must do some rearranging using some trig identities so that we have integrals that we can work with. We will need to remember the following identities as we work through this problem:

`tan^2(x)=sec^2(x)-1`

`tan(x)=(sin(x))/(cos(x))`

So, to begin, we will split up the `tan^3` like so:

`int tan^3(3x)dx=int tan(3x)*tan^2(3x)dx`

Next, we will use the first identity above to rewrite tan^2:

`int tan(3x)*tan^2(3x)dx=int tan(3x)(sec^2(3x)-1)dx`

We will then distribute and separate it into two integrals:

`int tan(3x)sec^2(3x)-tan(3x)dx=`

`= int tan(3x)sec^2(3x)dx-int tan(3x)dx`

Now we can use the second trig identity above to rewrite the second integrand:

`int tan(3x)sec^2(3x)dx-int (sin(3x))/(cos(3x))dx`

We can now use u-substitution on each integral. We will also need the chain rule on the 3x when finding du. So the u-substitution for the first integral is:

`u=tan(3x)`

`du=sec^2(3x)*3dx`

`1/3 du=sec^2(3x)dx`

For the second integral, we will use a **v** for our u-substitution so as not to confuse the two as we work through them. So the "v"-substitution for the second integral is:

`v=cos(3x)`

`dv=-sin(3x)*3dx`

`-1/3dv=sin(3x)dx`

Now, we will substitute these back into the integrals and integrate, as shown below:

`1/3 int udu-(-1/3)int (1/v)dv=(1/3)(u^2/2)+(1/3)ln|v|+C`

We can now substitute back in for u and v to find our final indefinite integral:

`int tan^3(3x)dx=1/6tan^2(3x)+1/3ln|cos(3x)|+C`