Integral of ctg(2x-1) dx

1. Rewrite the integrand:

   $$\cot{\left(2 x - 1 \right)} = \frac{\cos{\left(2 x - 1 \right)}}{\sin{\left(2 x - 1 \right)}}$$

2. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sin{\left(2 x - 1 \right)}$.

      Then let $du = 2 \cos{\left(2 x - 1 \right)} dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{1}{4 u}\, du$$

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

         $$\int \frac{1}{2 u}\, du = \frac{\int \frac{1}{u}\, du}{2}$$

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

         So, the result is: $\frac{\log{\left(u \right)}}{2}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(\sin{\left(2 x - 1 \right)} \right)}}{2}$$

   Method #2

   1. Let $u = 2 x - 1$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{\cos{\left(u \right)}}{4 \sin{\left(u \right)}}\, du$$

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

         $$\int \frac{\cos{\left(u \right)}}{2 \sin{\left(u \right)}}\, du = \frac{\int \frac{\cos{\left(u \right)}}{\sin{\left(u \right)}}\, du}{2}$$

         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: $\frac{\log{\left(\sin{\left(u \right)} \right)}}{2}$

      Now substitute $u$ back in:

      $$\frac{\log{\left(\sin{\left(2 x - 1 \right)} \right)}}{2}$$

3. Now simplify:

   $$\frac{\log{\left(\sin{\left(2 x - 1 \right)} \right)}}{2}$$

4. Add the constant of integration:

   $$\frac{\log{\left(\sin{\left(2 x - 1 \right)} \right)}}{2}+ \mathrm{constant}$$

The answer is: