Integral de $$$\sin^{4}{\left(\theta \right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \sin^{4}{\left(\theta \right)}\, d\theta$$$.
Solución
Aplica la fórmula de reducción de potencia $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ con $$$\alpha=\theta$$$:
$${\color{red}{\int{\sin^{4}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\left(- \frac{\cos{\left(2 \theta \right)}}{2} + \frac{\cos{\left(4 \theta \right)}}{8} + \frac{3}{8}\right)d \theta}}}$$
Aplica la regla del factor constante $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ con $$$c=\frac{1}{8}$$$ y $$$f{\left(\theta \right)} = - 4 \cos{\left(2 \theta \right)} + \cos{\left(4 \theta \right)} + 3$$$:
$${\color{red}{\int{\left(- \frac{\cos{\left(2 \theta \right)}}{2} + \frac{\cos{\left(4 \theta \right)}}{8} + \frac{3}{8}\right)d \theta}}} = {\color{red}{\left(\frac{\int{\left(- 4 \cos{\left(2 \theta \right)} + \cos{\left(4 \theta \right)} + 3\right)d \theta}}{8}\right)}}$$
Integra término a término:
$$\frac{{\color{red}{\int{\left(- 4 \cos{\left(2 \theta \right)} + \cos{\left(4 \theta \right)} + 3\right)d \theta}}}}{8} = \frac{{\color{red}{\left(\int{3 d \theta} - \int{4 \cos{\left(2 \theta \right)} d \theta} + \int{\cos{\left(4 \theta \right)} d \theta}\right)}}}{8}$$
Aplica la regla de la constante $$$\int c\, d\theta = c \theta$$$ con $$$c=3$$$:
$$- \frac{\int{4 \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\int{3 d \theta}}}}{8} = - \frac{\int{4 \cos{\left(2 \theta \right)} d \theta}}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} + \frac{{\color{red}{\left(3 \theta\right)}}}{8}$$
Aplica la regla del factor constante $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ con $$$c=4$$$ y $$$f{\left(\theta \right)} = \cos{\left(2 \theta \right)}$$$:
$$\frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{{\color{red}{\int{4 \cos{\left(2 \theta \right)} d \theta}}}}{8} = \frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{{\color{red}{\left(4 \int{\cos{\left(2 \theta \right)} d \theta}\right)}}}{8}$$
Sea $$$u=2 \theta$$$.
Entonces $$$du=\left(2 \theta\right)^{\prime }d\theta = 2 d\theta$$$ (los pasos pueden verse »), y obtenemos que $$$d\theta = \frac{du}{2}$$$.
La integral puede reescribirse como
$$\frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{{\color{red}{\int{\cos{\left(2 \theta \right)} d \theta}}}}{2} = \frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$
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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$
La integral del coseno es $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$
Recordemos que $$$u=2 \theta$$$:
$$\frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{3 \theta}{8} + \frac{\int{\cos{\left(4 \theta \right)} d \theta}}{8} - \frac{\sin{\left({\color{red}{\left(2 \theta\right)}} \right)}}{4}$$
Sea $$$u=4 \theta$$$.
Entonces $$$du=\left(4 \theta\right)^{\prime }d\theta = 4 d\theta$$$ (los pasos pueden verse »), y obtenemos que $$$d\theta = \frac{du}{4}$$$.
La integral puede reescribirse como
$$\frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(4 \theta \right)} d \theta}}}}{8} = \frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$
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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$
La integral del coseno es $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$
Recordemos que $$$u=4 \theta$$$:
$$\frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{\sin{\left({\color{red}{\left(4 \theta\right)}} \right)}}{32}$$
Por lo tanto,
$$\int{\sin^{4}{\left(\theta \right)} d \theta} = \frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{\sin{\left(4 \theta \right)}}{32}$$
Añade la constante de integración:
$$\int{\sin^{4}{\left(\theta \right)} d \theta} = \frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{\sin{\left(4 \theta \right)}}{32}+C$$
Respuesta
$$$\int \sin^{4}{\left(\theta \right)}\, d\theta = \left(\frac{3 \theta}{8} - \frac{\sin{\left(2 \theta \right)}}{4} + \frac{\sin{\left(4 \theta \right)}}{32}\right) + C$$$A