Integral de $$$\frac{8 \sin{\left(2 x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}}$$$

Halla $$$\int \frac{8 \sin{\left(2 x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}}\, dx$$$.

Reescribe el integrando:

$${\color{red}{\int{\frac{8 \sin{\left(2 x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}} d x}}} = {\color{red}{\int{16 \cos^{2}{\left(x \right)} d x}}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=16$$$ y $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$:

$${\color{red}{\int{16 \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\left(16 \int{\cos^{2}{\left(x \right)} d x}\right)}}$$

Aplica la fórmula de reducción de potencia $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ con $$$\alpha=x$$$:

$$16 {\color{red}{\int{\cos^{2}{\left(x \right)} d x}}} = 16 {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}$$

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

$$16 {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}} = 16 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}$$

Integra término a término:

$$8 {\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}} = 8 {\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

$$8 \int{\cos{\left(2 x \right)} d x} + 8 {\color{red}{\int{1 d x}}} = 8 \int{\cos{\left(2 x \right)} d x} + 8 {\color{red}{x}}$$

Sea $$$u=2 x$$$.

Entonces $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{2}$$$.

La integral se convierte en

$$8 x + 8 {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = 8 x + 8 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$

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

$$8 x + 8 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = 8 x + 8 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$

La integral del coseno es $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$8 x + 4 {\color{red}{\int{\cos{\left(u \right)} d u}}} = 8 x + 4 {\color{red}{\sin{\left(u \right)}}}$$

Recordemos que $$$u=2 x$$$:

$$8 x + 4 \sin{\left({\color{red}{u}} \right)} = 8 x + 4 \sin{\left({\color{red}{\left(2 x\right)}} \right)}$$

Por lo tanto,

$$\int{\frac{8 \sin{\left(2 x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}} d x} = 8 x + 4 \sin{\left(2 x \right)}$$

Añade la constante de integración:

$$\int{\frac{8 \sin{\left(2 x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}} d x} = 8 x + 4 \sin{\left(2 x \right)}+C$$

$$$\int \frac{8 \sin{\left(2 x \right)} \cos{\left(x \right)}}{\sin{\left(x \right)}}\, dx = \left(8 x + 4 \sin{\left(2 x \right)}\right) + C$$$A