Integral de $$$\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2}\, dx$$$.
Solución
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 $$$f{\left(x \right)} = \sec^{2}{\left(\frac{x}{2} \right)}$$$:
$${\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} d x}}} = {\color{red}{\left(\frac{\int{\sec^{2}{\left(\frac{x}{2} \right)} d x}}{2}\right)}}$$
Sea $$$u=\frac{x}{2}$$$.
Entonces $$$du=\left(\frac{x}{2}\right)^{\prime }dx = \frac{dx}{2}$$$ (los pasos pueden verse »), y obtenemos que $$$dx = 2 du$$$.
Por lo tanto,
$$\frac{{\color{red}{\int{\sec^{2}{\left(\frac{x}{2} \right)} d x}}}}{2} = \frac{{\color{red}{\int{2 \sec^{2}{\left(u \right)} d u}}}}{2}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=2$$$ y $$$f{\left(u \right)} = \sec^{2}{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{2 \sec^{2}{\left(u \right)} d u}}}}{2} = \frac{{\color{red}{\left(2 \int{\sec^{2}{\left(u \right)} d u}\right)}}}{2}$$
La integral de $$$\sec^{2}{\left(u \right)}$$$ es $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$:
$${\color{red}{\int{\sec^{2}{\left(u \right)} d u}}} = {\color{red}{\tan{\left(u \right)}}}$$
Recordemos que $$$u=\frac{x}{2}$$$:
$$\tan{\left({\color{red}{u}} \right)} = \tan{\left({\color{red}{\left(\frac{x}{2}\right)}} \right)}$$
Por lo tanto,
$$\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} d x} = \tan{\left(\frac{x}{2} \right)}$$
Añade la constante de integración:
$$\int{\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} d x} = \tan{\left(\frac{x}{2} \right)}+C$$
Respuesta
$$$\int \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2}\, dx = \tan{\left(\frac{x}{2} \right)} + C$$$A