Integral of sin(2*x)/cos(2*x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x$.

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

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

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

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

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

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

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

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

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

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

      Now substitute $u$ back in:

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

   Method #2

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

      $$\int \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\cos{\left(2 x \right)}}\, dx = 2 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\cos{\left(2 x \right)}}\, dx$$

      1. Rewrite the integrand:

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

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

         Then let $du = - 4 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $- \frac{du}{4}$:

         $$\int \left(- \frac{1}{4 u}\right)\, du$$

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

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

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

            So, the result is: $- \frac{\log{\left(u \right)}}{4}$

         Now substitute $u$ back in:

         $$- \frac{\log{\left(2 \cos^{2}{\left(x \right)} - 1 \right)}}{4}$$

      So, the result is: $- \frac{\log{\left(2 \cos^{2}{\left(x \right)} - 1 \right)}}{2}$

2. Add the constant of integration:

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

The answer is:

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