Integral of ctg3x dx

1. Rewrite the integrand:

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

2. There are multiple ways to do this integral.

   Method #1

   1. 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:

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

   Method #2

   1. Let $u = 3 x$.

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

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

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

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

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

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

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

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

            Now substitute $u$ back in:

            $$\log{\left(\sin{\left(u \right)} \right)}$$

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

      Now substitute $u$ back in:

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

3. Add the constant of integration:

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

The answer is: