Integral de $$$\cos{\left(2 x \right)} \tan{\left(x \right)}$$$

Halla $$$\int \cos{\left(2 x \right)} \tan{\left(x \right)}\, dx$$$.

Reescribe el integrando utilizando la fórmula del ángulo doble del coseno $$$\cos{\left(2 x \right)}=2 \cos^{2}{\left(x \right)} - 1$$$:

$${\color{red}{\int{\cos{\left(2 x \right)} \tan{\left(x \right)} d x}}} = {\color{red}{\int{\left(2 \cos^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} d x}}}$$

Reescribir:

$${\color{red}{\int{\left(2 \cos^{2}{\left(x \right)} - 1\right) \tan{\left(x \right)} d x}}} = {\color{red}{\int{\left(2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} - \tan{\left(x \right)}\right)d x}}}$$

Integra término a término:

$${\color{red}{\int{\left(2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} - \tan{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - \int{\tan{\left(x \right)} d x}\right)}}$$

Reescribe la tangente como $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

$$\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - {\color{red}{\int{\tan{\left(x \right)} d x}}} = \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$

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

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

Por lo tanto,

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

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

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

La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x} = \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)} + \int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=2$$$ y $$$f{\left(x \right)} = \cos^{2}{\left(x \right)} \tan{\left(x \right)}$$$:

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{2 \cos^{2}{\left(x \right)} \tan{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\left(2 \int{\cos^{2}{\left(x \right)} \tan{\left(x \right)} d x}\right)}}$$

Reescribe el integrando:

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\int{\cos^{2}{\left(x \right)} \tan{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + 2 {\color{red}{\int{\sin{\left(x \right)} \cos{\left(x \right)} d x}}}$$

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

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

La integral se convierte en

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

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=1$$$:

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

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

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{u}}^{2} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\sin{\left(x \right)}}}^{2}$$

Por lo tanto,

$$\int{\cos{\left(2 x \right)} \tan{\left(x \right)} d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \sin^{2}{\left(x \right)}$$

Añade la constante de integración:

$$\int{\cos{\left(2 x \right)} \tan{\left(x \right)} d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \sin^{2}{\left(x \right)}+C$$

$$$\int \cos{\left(2 x \right)} \tan{\left(x \right)}\, dx = \left(\ln\left(\left|{\cos{\left(x \right)}}\right|\right) + \sin^{2}{\left(x \right)}\right) + C$$$A