Integral de $$$- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}$$$ con respecto a $$$x$$$

Halla $$$\int \left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=- 9 i n t \sec{\left(2 \right)}$$$ y $$$f{\left(x \right)} = x \sin{\left(3 x \right)}$$$:

$${\color{red}{\int{\left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)d x}}} = {\color{red}{\left(- 9 i n t \sec{\left(2 \right)} \int{x \sin{\left(3 x \right)} d x}\right)}}$$

Para la integral $$$\int{x \sin{\left(3 x \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=x$$$ y $$$\operatorname{dv}=\sin{\left(3 x \right)} dx$$$.

Entonces $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{\sin{\left(3 x \right)} d x}=- \frac{\cos{\left(3 x \right)}}{3}$$$ (los pasos pueden verse »).

La integral puede reescribirse como

$$- 9 i n t \sec{\left(2 \right)} {\color{red}{\int{x \sin{\left(3 x \right)} d x}}}=- 9 i n t \sec{\left(2 \right)} {\color{red}{\left(x \cdot \left(- \frac{\cos{\left(3 x \right)}}{3}\right)-\int{\left(- \frac{\cos{\left(3 x \right)}}{3}\right) \cdot 1 d x}\right)}}=- 9 i n t \sec{\left(2 \right)} {\color{red}{\left(- \frac{x \cos{\left(3 x \right)}}{3} - \int{\left(- \frac{\cos{\left(3 x \right)}}{3}\right)d x}\right)}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=- \frac{1}{3}$$$ y $$$f{\left(x \right)} = \cos{\left(3 x \right)}$$$:

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} - {\color{red}{\int{\left(- \frac{\cos{\left(3 x \right)}}{3}\right)d x}}}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} - {\color{red}{\left(- \frac{\int{\cos{\left(3 x \right)} d x}}{3}\right)}}\right)$$

Sea $$$u=3 x$$$.

Entonces $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{3}$$$.

Entonces,

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\cos{\left(3 x \right)} d x}}}}{3}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{3}\right)$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{3}$$$ y $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{3}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}}}{3}\right)$$

La integral del coseno es $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{9}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\sin{\left(u \right)}}}}{9}\right)$$

Recordemos que $$$u=3 x$$$:

$$- 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left({\color{red}{u}} \right)}}{9}\right) = - 9 i n t \sec{\left(2 \right)} \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{9}\right)$$

Por lo tanto,

$$\int{\left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)d x} = - 9 i n t \left(- \frac{x \cos{\left(3 x \right)}}{3} + \frac{\sin{\left(3 x \right)}}{9}\right) \sec{\left(2 \right)}$$

Simplificar:

$$\int{\left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)d x} = i n t \left(3 x \cos{\left(3 x \right)} - \sin{\left(3 x \right)}\right) \sec{\left(2 \right)}$$

Añade la constante de integración:

$$\int{\left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)d x} = i n t \left(3 x \cos{\left(3 x \right)} - \sin{\left(3 x \right)}\right) \sec{\left(2 \right)}+C$$

$$$\int \left(- 9 i n t x \sin{\left(3 x \right)} \sec{\left(2 \right)}\right)\, dx = i n t \left(3 x \cos{\left(3 x \right)} - \sin{\left(3 x \right)}\right) \sec{\left(2 \right)} + C$$$A