Integral de $$$\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}}$$$
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Tu entrada
Halla $$$\int \frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}}\, dt$$$.
Solución
Reescribe el integrando:
$${\color{red}{\int{\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}} d t}}} = {\color{red}{\int{\frac{\sec^{2}{\left(t \right)}}{\tan{\left(t \right)}} d t}}}$$
Sea $$$u=\tan{\left(t \right)}$$$.
Entonces $$$du=\left(\tan{\left(t \right)}\right)^{\prime }dt = \sec^{2}{\left(t \right)} dt$$$ (los pasos pueden verse »), y obtenemos que $$$\sec^{2}{\left(t \right)} dt = du$$$.
La integral puede reescribirse como
$${\color{red}{\int{\frac{\sec^{2}{\left(t \right)}}{\tan{\left(t \right)}} d t}}} = {\color{red}{\int{\frac{1}{u} d u}}}$$
La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$${\color{red}{\int{\frac{1}{u} d u}}} = {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Recordemos que $$$u=\tan{\left(t \right)}$$$:
$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\tan{\left(t \right)}}}}\right| \right)}$$
Por lo tanto,
$$\int{\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}} d t} = \ln{\left(\left|{\tan{\left(t \right)}}\right| \right)}$$
Añade la constante de integración:
$$\int{\frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}} d t} = \ln{\left(\left|{\tan{\left(t \right)}}\right| \right)}+C$$
Respuesta
$$$\int \frac{1}{\cos^{2}{\left(t \right)} \tan{\left(t \right)}}\, dt = \ln\left(\left|{\tan{\left(t \right)}}\right|\right) + C$$$A