Integral de $$$\sin^{3}{\left(2 x \right)}$$$

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

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,

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

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 $$$f{\left(u \right)} = \sin^{3}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\sin^{3}{\left(u \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\sin^{3}{\left(u \right)} d u}}{2}\right)}}$$

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

$$\frac{{\color{red}{\int{\sin^{3}{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\int{\left(1 - \cos^{2}{\left(u \right)}\right) \sin{\left(u \right)} d u}}}}{2}$$

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

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

Por lo tanto,

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

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

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dv = c v$$$ con $$$c=1$$$:

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

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

$$- \frac{v}{2} + \frac{{\color{red}{\int{v^{2} d v}}}}{2}=- \frac{v}{2} + \frac{{\color{red}{\frac{v^{1 + 2}}{1 + 2}}}}{2}=- \frac{v}{2} + \frac{{\color{red}{\left(\frac{v^{3}}{3}\right)}}}{2}$$

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

$$- \frac{{\color{red}{v}}}{2} + \frac{{\color{red}{v}}^{3}}{6} = - \frac{{\color{red}{\cos{\left(u \right)}}}}{2} + \frac{{\color{red}{\cos{\left(u \right)}}}^{3}}{6}$$

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

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{2} + \frac{\cos^{3}{\left({\color{red}{u}} \right)}}{6} = - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{2} + \frac{\cos^{3}{\left({\color{red}{\left(2 x\right)}} \right)}}{6}$$

Por lo tanto,

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

Simplificar:

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

Añade la constante de integración:

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

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