Integral of $$$\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}$$$

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

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

Then $$$du=\left(\sin{\left(2 x \right)}\right)^{\prime }dx = 2 \cos{\left(2 x \right)} dx$$$ (steps can be seen »), and we have that $$$\cos{\left(2 x \right)} dx = \frac{du}{2}$$$.

The integral can be rewritten as

$${\color{red}{\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x}}} = {\color{red}{\int{\frac{1}{2 u} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \frac{1}{u}$$$:

$${\color{red}{\int{\frac{1}{2 u} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{2}\right)}}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

Recall that $$$u=\sin{\left(2 x \right)}$$$:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{{\color{red}{\sin{\left(2 x \right)}}}}\right| \right)}}{2}$$

Therefore,

$$\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{2}$$

Add the constant of integration:

$$\int{\frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}} d x} = \frac{\ln{\left(\left|{\sin{\left(2 x \right)}}\right| \right)}}{2}+C$$

$$$\int \frac{\cos{\left(2 x \right)}}{\sin{\left(2 x \right)}}\, dx = \frac{\ln\left(\left|{\sin{\left(2 x \right)}}\right|\right)}{2} + C$$$A