Integral de $$$\frac{\sin{\left(2 x \right)}}{3} - \cos^{4}{\left(x \right)}$$$

Halla $$$\int \left(\frac{\sin{\left(2 x \right)}}{3} - \cos^{4}{\left(x \right)}\right)\, dx$$$.

Integra término a término:

$${\color{red}{\int{\left(\frac{\sin{\left(2 x \right)}}{3} - \cos^{4}{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{\frac{\sin{\left(2 x \right)}}{3} d x} - \int{\cos^{4}{\left(x \right)} d x}\right)}}$$

Aplica la fórmula de reducción de potencia $$$\cos^{4}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ con $$$\alpha=x$$$:

$$\int{\frac{\sin{\left(2 x \right)}}{3} d x} - {\color{red}{\int{\cos^{4}{\left(x \right)} d x}}} = \int{\frac{\sin{\left(2 x \right)}}{3} d x} - {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{8} + \frac{3}{8}\right)d x}}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{8}$$$ y $$$f{\left(x \right)} = 4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3$$$:

$$\int{\frac{\sin{\left(2 x \right)}}{3} d x} - {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{8} + \frac{3}{8}\right)d x}}} = \int{\frac{\sin{\left(2 x \right)}}{3} d x} - {\color{red}{\left(\frac{\int{\left(4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3\right)d x}}{8}\right)}}$$

Integra término a término:

$$\int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{{\color{red}{\int{\left(4 \cos{\left(2 x \right)} + \cos{\left(4 x \right)} + 3\right)d x}}}}{8} = \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{{\color{red}{\left(\int{3 d x} + \int{4 \cos{\left(2 x \right)} d x} + \int{\cos{\left(4 x \right)} d x}\right)}}}{8}$$

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=3$$$:

$$\int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{4 \cos{\left(2 x \right)} d x}}{8} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{3 d x}}}}{8} = \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{4 \cos{\left(2 x \right)} d x}}{8} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(3 x\right)}}}{8}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=4$$$ y $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

$$- \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{4 \cos{\left(2 x \right)} d x}}}}{8} = - \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(4 \int{\cos{\left(2 x \right)} d x}\right)}}}{8}$$

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

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

Entonces,

$$- \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{2} = - \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$

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

$$- \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = - \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$

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

$$- \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = - \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

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

$$- \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = - \frac{3 x}{8} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Sea $$$u=4 x$$$.

Entonces $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{4}$$$.

Entonces,

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{8} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{4}$$$ y $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$

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

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$

Recordemos que $$$u=4 x$$$:

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\sin{\left({\color{red}{u}} \right)}}{32} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} + \int{\frac{\sin{\left(2 x \right)}}{3} d x} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{32}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{3}$$$ y $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} + {\color{red}{\int{\frac{\sin{\left(2 x \right)}}{3} d x}}} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} + {\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{3}\right)}}$$

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 puede reescribirse como

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} + \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{3} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{3}$$

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

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{3} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{3}$$

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

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{6} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{6}$$

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

$$- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} - \frac{\cos{\left({\color{red}{u}} \right)}}{6} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{6}$$

Por lo tanto,

$$\int{\left(\frac{\sin{\left(2 x \right)}}{3} - \cos^{4}{\left(x \right)}\right)d x} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} - \frac{\cos{\left(2 x \right)}}{6}$$

Añade la constante de integración:

$$\int{\left(\frac{\sin{\left(2 x \right)}}{3} - \cos^{4}{\left(x \right)}\right)d x} = - \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} - \frac{\cos{\left(2 x \right)}}{6}+C$$

$$$\int \left(\frac{\sin{\left(2 x \right)}}{3} - \cos^{4}{\left(x \right)}\right)\, dx = \left(- \frac{3 x}{8} - \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{32} - \frac{\cos{\left(2 x \right)}}{6}\right) + C$$$A