Integral de $$$\frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}$$$

Halla $$$\int \frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx$$$.

Multiplica el numerador y el denominador por un seno y expresa todo lo demás en términos del coseno, usando la fórmula $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$ con $$$\alpha=x$$$:

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

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

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

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = \frac{1}{u^{2} \left(1 - u^{2}\right)}$$$:

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

Realizar la descomposición en fracciones parciales (los pasos pueden verse »):

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

Integra término a término:

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

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=-2$$$:

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \frac{1}{u + 1}$$$:

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

Sea $$$v=u + 1$$$.

Entonces $$$dv=\left(u + 1\right)^{\prime }du = 1 du$$$ (los pasos pueden verse »), y obtenemos que $$$du = dv$$$.

Por lo tanto,

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

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

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

Recordemos que $$$v=u + 1$$$:

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \frac{1}{u - 1}$$$:

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

Sea $$$v=u - 1$$$.

Entonces $$$dv=\left(u - 1\right)^{\prime }du = 1 du$$$ (los pasos pueden verse »), y obtenemos que $$$du = dv$$$.

Entonces,

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

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

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

Recordemos que $$$v=u - 1$$$:

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

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

$$\frac{\ln{\left(\left|{-1 + {\color{red}{u}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{u}}}\right| \right)}}{2} + {\color{red}{u}}^{-1} = \frac{\ln{\left(\left|{-1 + {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{\cos{\left(x \right)}}}}\right| \right)}}{2} + {\color{red}{\cos{\left(x \right)}}}^{-1}$$

Por lo tanto,

$$\int{\frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}} d x} = \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} + \frac{1}{\cos{\left(x \right)}}$$

Añade la constante de integración:

$$\int{\frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}} d x} = \frac{\ln{\left(\left|{\cos{\left(x \right)} - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{\cos{\left(x \right)} + 1}\right| \right)}}{2} + \frac{1}{\cos{\left(x \right)}}+C$$

$$$\int \frac{1}{\sin{\left(x \right)} \cos^{2}{\left(x \right)}}\, dx = \left(\frac{\ln\left(\left|{\cos{\left(x \right)} - 1}\right|\right)}{2} - \frac{\ln\left(\left|{\cos{\left(x \right)} + 1}\right|\right)}{2} + \frac{1}{\cos{\left(x \right)}}\right) + C$$$A