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Integral de $$$\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}\, dx$$$.
Solución
Sea $$$u=8 x$$$.
Entonces $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{8}$$$.
Por lo tanto,
$${\color{red}{\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x}}} = {\color{red}{\int{\frac{\sin{\left(2 u \right)}}{8 \cos{\left(u \right)}} d u}}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{8}$$$ y $$$f{\left(u \right)} = \frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}}$$$:
$${\color{red}{\int{\frac{\sin{\left(2 u \right)}}{8 \cos{\left(u \right)}} d u}}} = {\color{red}{\left(\frac{\int{\frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}} d u}}{8}\right)}}$$
Reescribe el integrando:
$$\frac{{\color{red}{\int{\frac{\sin{\left(2 u \right)}}{\cos{\left(u \right)}} d u}}}}{8} = \frac{{\color{red}{\int{2 \sin{\left(u \right)} d u}}}}{8}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=2$$$ y $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{2 \sin{\left(u \right)} d u}}}}{8} = \frac{{\color{red}{\left(2 \int{\sin{\left(u \right)} d u}\right)}}}{8}$$
La integral del seno es $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}$$
Recordemos que $$$u=8 x$$$:
$$- \frac{\cos{\left({\color{red}{u}} \right)}}{4} = - \frac{\cos{\left({\color{red}{\left(8 x\right)}} \right)}}{4}$$
Por lo tanto,
$$\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x} = - \frac{\cos{\left(8 x \right)}}{4}$$
Añade la constante de integración:
$$\int{\frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}} d x} = - \frac{\cos{\left(8 x \right)}}{4}+C$$
Respuesta
$$$\int \frac{\sin{\left(16 x \right)}}{\cos{\left(8 x \right)}}\, dx = - \frac{\cos{\left(8 x \right)}}{4} + C$$$A