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

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

Let $u=2 x$.

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

Therefore,

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

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

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

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

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

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 $\sin{\left(u \right)} du = - dv$.

Thus,

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

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

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

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

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

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

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

Recall that $u=2 x$:

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

Therefore,

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

Add the constant of integration:

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

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