Integral of tg(3x)dx dx

1. Let $u = 3 x$.

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

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

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

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

      1. Rewrite the integrand:

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

      2. 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:

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

2. Add the constant of integration:

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

The answer is: