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

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

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

$${\color{red}{\int{\frac{\tan{\left(2 x \right)}}{2} d x}}} = {\color{red}{\left(\frac{\int{\tan{\left(2 x \right)} d x}}{2}\right)}}$$

Let $$$u=2 x$$$.

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

So,

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

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 $$$f{\left(u \right)} = \tan{\left(u \right)}$$$:

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

Rewrite the tangent as $$$\tan\left( u \right)=\frac{\sin\left( u \right)}{\cos\left( u \right)}$$$:

$$\frac{{\color{red}{\int{\tan{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{\cos{\left(u \right)}} d u}}}}{4}$$

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

Then $$$dv=\left(\cos{\left(u \right)}\right)^{\prime }du = - \sin{\left(u \right)} du$$$ (steps can be seen »), and we have that $$$\sin{\left(u \right)} du = - dv$$$.

The integral becomes

$$\frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{\cos{\left(u \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\left(- \frac{1}{v}\right)d v}}}}{4}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=-1$$$ and $$$f{\left(v \right)} = \frac{1}{v}$$$:

$$\frac{{\color{red}{\int{\left(- \frac{1}{v}\right)d v}}}}{4} = \frac{{\color{red}{\left(- \int{\frac{1}{v} d v}\right)}}}{4}$$

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

$$- \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = - \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$

Recall that $$$v=\cos{\left(u \right)}$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = - \frac{\ln{\left(\left|{{\color{red}{\cos{\left(u \right)}}}}\right| \right)}}{4}$$

Recall that $$$u=2 x$$$:

$$- \frac{\ln{\left(\left|{\cos{\left({\color{red}{u}} \right)}}\right| \right)}}{4} = - \frac{\ln{\left(\left|{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}\right| \right)}}{4}$$

Therefore,

$$\int{\frac{\tan{\left(2 x \right)}}{2} d x} = - \frac{\ln{\left(\left|{\cos{\left(2 x \right)}}\right| \right)}}{4}$$

Add the constant of integration:

$$\int{\frac{\tan{\left(2 x \right)}}{2} d x} = - \frac{\ln{\left(\left|{\cos{\left(2 x \right)}}\right| \right)}}{4}+C$$

$$$\int \frac{\tan{\left(2 x \right)}}{2}\, dx = - \frac{\ln\left(\left|{\cos{\left(2 x \right)}}\right|\right)}{4} + C$$$A