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