Integral of tg(3x)+ctg(3x) dx

1. Integrate term-by-term:

   1. Rewrite the integrand:

      $$\tan{\left(3 x \right)} = \frac{\sin{\left(3 x \right)}}{\cos{\left(3 x \right)}}$$

   2. There are multiple ways to do this integral.

      Method #1

      1. Let $u = \cos{\left(3 x \right)}$.

         Then let $du = - 3 \sin{\left(3 x \right)} dx$ and substitute $- \frac{du}{3}$:

         $$\int \frac{1}{9 u}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \left(- \frac{1}{3 u}\right)\, du = - \frac{\int \frac{1}{u}\, du}{3}$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            So, the result is: $- \frac{\log{\left(u \right)}}{3}$

         Now substitute $u$ back in:

         $$- \frac{\log{\left(\cos{\left(3 x \right)} \right)}}{3}$$

      Method #2

      1. Let $u = 3 x$.

         Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

         $$\int \frac{\sin{\left(u \right)}}{9 \cos{\left(u \right)}}\, du$$

         1. The integral of a constant times a function is the constant times the integral of the function:

            $$\int \frac{\sin{\left(u \right)}}{3 \cos{\left(u \right)}}\, du = \frac{\int \frac{\sin{\left(u \right)}}{\cos{\left(u \right)}}\, du}{3}$$

            1. Let $u = \cos{\left(u \right)}$.

               Then let $du = - \sin{\left(u \right)} du$ and substitute $- du$:

               $$\int \frac{1}{u}\, du$$

               1. The integral of a constant times a function is the constant times the integral of the function:

                  $$\int \left(- \frac{1}{u}\right)\, du = - \int \frac{1}{u}\, du$$

                  1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

                  So, the result is: $- \log{\left(u \right)}$

               Now substitute $u$ back in:

               $$- \log{\left(\cos{\left(u \right)} \right)}$$

            So, the result is: $- \frac{\log{\left(\cos{\left(u \right)} \right)}}{3}$

         Now substitute $u$ back in:

         $$- \frac{\log{\left(\cos{\left(3 x \right)} \right)}}{3}$$

   1. Rewrite the integrand:

      $$\cot{\left(3 x \right)} = \frac{\cos{\left(3 x \right)}}{\sin{\left(3 x \right)}}$$

   2. Let $u = \sin{\left(3 x \right)}$.

      Then let $du = 3 \cos{\left(3 x \right)} dx$ and substitute $\frac{du}{3}$:

      $$\int \frac{1}{9 u}\, du$$

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \frac{1}{3 u}\, du = \frac{\int \frac{1}{u}\, du}{3}$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         So, the result is: $\frac{\log{\left(u \right)}}{3}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(\sin{\left(3 x \right)} \right)}}{3}$$

   The result is: $\frac{\log{\left(\sin{\left(3 x \right)} \right)}}{3} - \frac{\log{\left(\cos{\left(3 x \right)} \right)}}{3}$

2. Add the constant of integration:

   $$\frac{\log{\left(\sin{\left(3 x \right)} \right)}}{3} - \frac{\log{\left(\cos{\left(3 x \right)} \right)}}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{\log{\left(\sin{\left(3 x \right)} \right)}}{3} - \frac{\log{\left(\cos{\left(3 x \right)} \right)}}{3}+ \mathrm{constant}$$