Integral de $$$\tan{\left(y \right)}$$$

Halla $$$\int \tan{\left(y \right)}\, dy$$$.

Reescribe la tangente como $$$\tan\left(y\right)=\frac{\sin\left(y\right)}{\cos\left(y\right)}$$$:

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

Sea $$$u=\cos{\left(y \right)}$$$.

Entonces $$$du=\left(\cos{\left(y \right)}\right)^{\prime }dy = - \sin{\left(y \right)} dy$$$ (los pasos pueden verse »), y obtenemos que $$$\sin{\left(y \right)} dy = - du$$$.

Entonces,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:

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

La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

Recordemos que $$$u=\cos{\left(y \right)}$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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