Integral de $$$- x \sin{\left(x \right)} \tan{\left(1 \right)}$$$

Halla $$$\int \left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)\, dx$$$.

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

$${\color{red}{\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x}}} = {\color{red}{\left(- \tan{\left(1 \right)} \int{x \sin{\left(x \right)} d x}\right)}}$$

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

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

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

La integral se convierte en

$$- \tan{\left(1 \right)} {\color{red}{\int{x \sin{\left(x \right)} d x}}}=- \tan{\left(1 \right)} {\color{red}{\left(x \cdot \left(- \cos{\left(x \right)}\right)-\int{\left(- \cos{\left(x \right)}\right) \cdot 1 d x}\right)}}=- \tan{\left(1 \right)} {\color{red}{\left(- x \cos{\left(x \right)} - \int{\left(- \cos{\left(x \right)}\right)d x}\right)}}$$

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

$$- \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} - {\color{red}{\int{\left(- \cos{\left(x \right)}\right)d x}}}\right) = - \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} - {\color{red}{\left(- \int{\cos{\left(x \right)} d x}\right)}}\right)$$

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

$$- \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} + {\color{red}{\int{\cos{\left(x \right)} d x}}}\right) = - \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} + {\color{red}{\sin{\left(x \right)}}}\right)$$

Por lo tanto,

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = - \left(- x \cos{\left(x \right)} + \sin{\left(x \right)}\right) \tan{\left(1 \right)}$$

Simplificar:

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)}$$

Añade la constante de integración:

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)}+C$$

$$$\int \left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)\, dx = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)} + C$$$A