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