Integral de $$$\sin{\left(\ln\left(2 x\right) \right)}$$$

Halla $$$\int \sin{\left(\ln\left(2 x\right) \right)}\, dx$$$.

Sea $$$u=2 x$$$.

Entonces $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{2}$$$.

Entonces,

$${\color{red}{\int{\sin{\left(\ln{\left(2 x \right)} \right)} d x}}} = {\color{red}{\int{\frac{\sin{\left(\ln{\left(u \right)} \right)}}{2} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \sin{\left(\ln{\left(u \right)} \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(\ln{\left(u \right)} \right)}}{2} d u}}} = {\color{red}{\left(\frac{\int{\sin{\left(\ln{\left(u \right)} \right)} d u}}{2}\right)}}$$

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

Sean $$$\operatorname{\kappa}=\sin{\left(\ln{\left(u \right)} \right)}$$$ y $$$\operatorname{dv}=du$$$.

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

La integral puede reescribirse como

$$\frac{{\color{red}{\int{\sin{\left(\ln{\left(u \right)} \right)} d u}}}}{2}=\frac{{\color{red}{\left(\sin{\left(\ln{\left(u \right)} \right)} \cdot u-\int{u \cdot \frac{\cos{\left(\ln{\left(u \right)} \right)}}{u} d u}\right)}}}{2}=\frac{{\color{red}{\left(u \sin{\left(\ln{\left(u \right)} \right)} - \int{\cos{\left(\ln{\left(u \right)} \right)} d u}\right)}}}{2}$$

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

Sean $$$\operatorname{\kappa}=\cos{\left(\ln{\left(u \right)} \right)}$$$ y $$$\operatorname{dv}=du$$$.

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

Entonces,

$$\frac{u \sin{\left(\ln{\left(u \right)} \right)}}{2} - \frac{{\color{red}{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}}}{2}=\frac{u \sin{\left(\ln{\left(u \right)} \right)}}{2} - \frac{{\color{red}{\left(\cos{\left(\ln{\left(u \right)} \right)} \cdot u-\int{u \cdot \left(- \frac{\sin{\left(\ln{\left(u \right)} \right)}}{u}\right) d u}\right)}}}{2}=\frac{u \sin{\left(\ln{\left(u \right)} \right)}}{2} - \frac{{\color{red}{\left(u \cos{\left(\ln{\left(u \right)} \right)} - \int{\left(- \sin{\left(\ln{\left(u \right)} \right)}\right)d u}\right)}}}{2}$$

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

$$\frac{u \sin{\left(\ln{\left(u \right)} \right)}}{2} - \frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} + \frac{{\color{red}{\int{\left(- \sin{\left(\ln{\left(u \right)} \right)}\right)d u}}}}{2} = \frac{u \sin{\left(\ln{\left(u \right)} \right)}}{2} - \frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} + \frac{{\color{red}{\left(- \int{\sin{\left(\ln{\left(u \right)} \right)} d u}\right)}}}{2}$$

Hemos llegado a una integral que ya hemos visto.

Así, hemos obtenido la siguiente ecuación simple con respecto a la integral:

$$\frac{\int{\sin{\left(\ln{\left(u \right)} \right)} d u}}{2} = \frac{u \sin{\left(\ln{\left(u \right)} \right)}}{2} - \frac{u \cos{\left(\ln{\left(u \right)} \right)}}{2} - \frac{\int{\sin{\left(\ln{\left(u \right)} \right)} d u}}{2}$$

Al resolverlo, obtenemos que

$$\int{\sin{\left(\ln{\left(u \right)} \right)} d u} = \frac{u \left(\sin{\left(\ln{\left(u \right)} \right)} - \cos{\left(\ln{\left(u \right)} \right)}\right)}{2}$$

Por lo tanto,

$$\frac{{\color{red}{\int{\sin{\left(\ln{\left(u \right)} \right)} d u}}}}{2} = \frac{{\color{red}{\left(\frac{u \left(\sin{\left(\ln{\left(u \right)} \right)} - \cos{\left(\ln{\left(u \right)} \right)}\right)}{2}\right)}}}{2}$$

Recordemos que $$$u=2 x$$$:

$$\frac{{\color{red}{u}} \left(\sin{\left(\ln{\left({\color{red}{u}} \right)} \right)} - \cos{\left(\ln{\left({\color{red}{u}} \right)} \right)}\right)}{4} = \frac{{\color{red}{\left(2 x\right)}} \left(\sin{\left(\ln{\left({\color{red}{\left(2 x\right)}} \right)} \right)} - \cos{\left(\ln{\left({\color{red}{\left(2 x\right)}} \right)} \right)}\right)}{4}$$

Por lo tanto,

$$\int{\sin{\left(\ln{\left(2 x \right)} \right)} d x} = \frac{x \left(\sin{\left(\ln{\left(2 x \right)} \right)} - \cos{\left(\ln{\left(2 x \right)} \right)}\right)}{2}$$

Simplificar:

$$\int{\sin{\left(\ln{\left(2 x \right)} \right)} d x} = - \frac{\sqrt{2} x \cos{\left(\ln{\left(x \right)} + \ln{\left(2 \right)} + \frac{\pi}{4} \right)}}{2}$$

Añade la constante de integración:

$$\int{\sin{\left(\ln{\left(2 x \right)} \right)} d x} = - \frac{\sqrt{2} x \cos{\left(\ln{\left(x \right)} + \ln{\left(2 \right)} + \frac{\pi}{4} \right)}}{2}+C$$

$$$\int \sin{\left(\ln\left(2 x\right) \right)}\, dx = - \frac{\sqrt{2} x \cos{\left(\ln\left(x\right) + \ln\left(2\right) + \frac{\pi}{4} \right)}}{2} + C$$$A