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

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

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

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

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

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

Por lo tanto,

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

Como el grado del numerador no es menor que el grado del denominador, realiza la división larga de polinomios (los pasos pueden verse »):

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

$$\int{\frac{1}{1 - u^{2}} d u} - {\color{red}{\int{1 d u}}} = \int{\frac{1}{1 - u^{2}} d u} - {\color{red}{u}}$$

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

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

Integra término a término:

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

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 $$$f{\left(u \right)} = \frac{1}{u + 1}$$$:

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

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,

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

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

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

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

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

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

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

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

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

La integral puede reescribirse como

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

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

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

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

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

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