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

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

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

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

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

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

Therefore,

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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