Integral de $$$\frac{1}{x \sec^{2}{\left(\ln\left(x\right) \right)}}$$$

Halla $$$\int \frac{1}{x \sec^{2}{\left(\ln\left(x\right) \right)}}\, dx$$$.

Sea $$$u=\ln{\left(x \right)}$$$.

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

Entonces,

$${\color{red}{\int{\frac{1}{x \sec^{2}{\left(\ln{\left(x \right)} \right)}} d x}}} = {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}$$

Aplica la fórmula de reducción de potencia $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ con $$$\alpha= u $$$:

$${\color{red}{\int{\cos^{2}{\left(u \right)} d u}}} = {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}$$

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

$${\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}} = {\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}$$

Integra término a término:

$$\frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{2} = \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{2}$$

Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:

$$\frac{\int{\cos{\left(2 u \right)} d u}}{2} + \frac{{\color{red}{\int{1 d u}}}}{2} = \frac{\int{\cos{\left(2 u \right)} d u}}{2} + \frac{{\color{red}{u}}}{2}$$

Sea $$$v=2 u$$$.

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

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

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

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

$$\frac{u}{2} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{4} = \frac{u}{2} + \frac{{\color{red}{\sin{\left(v \right)}}}}{4}$$

Recordemos que $$$v=2 u$$$:

$$\frac{u}{2} + \frac{\sin{\left({\color{red}{v}} \right)}}{4} = \frac{u}{2} + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{4}$$

Recordemos que $$$u=\ln{\left(x \right)}$$$:

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

Por lo tanto,

$$\int{\frac{1}{x \sec^{2}{\left(\ln{\left(x \right)} \right)}} d x} = \frac{\ln{\left(x \right)}}{2} + \frac{\sin{\left(2 \ln{\left(x \right)} \right)}}{4}$$

Añade la constante de integración:

$$\int{\frac{1}{x \sec^{2}{\left(\ln{\left(x \right)} \right)}} d x} = \frac{\ln{\left(x \right)}}{2} + \frac{\sin{\left(2 \ln{\left(x \right)} \right)}}{4}+C$$

$$$\int \frac{1}{x \sec^{2}{\left(\ln\left(x\right) \right)}}\, dx = \left(\frac{\ln\left(x\right)}{2} + \frac{\sin{\left(2 \ln\left(x\right) \right)}}{4}\right) + C$$$A