Integral de $$$- \tan{\left(x \right)} + \sec{\left(x \right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)\, dx$$$.
Solución
Integra término a término:
$${\color{red}{\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{\tan{\left(x \right)} d x} + \int{\sec{\left(x \right)} d x}\right)}}$$
Reescribe la tangente como $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:
$$\int{\sec{\left(x \right)} d x} - {\color{red}{\int{\tan{\left(x \right)} d x}}} = \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}}$$
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 $$$\sin{\left(x \right)} dx = - du$$$.
Entonces,
$$\int{\sec{\left(x \right)} d x} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} d x}}} = \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = \frac{1}{u}$$$:
$$\int{\sec{\left(x \right)} d x} - {\color{red}{\int{\left(- \frac{1}{u}\right)d u}}} = \int{\sec{\left(x \right)} d x} - {\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}$$
La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\int{\sec{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{u} d u}}} = \int{\sec{\left(x \right)} d x} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Recordemos que $$$u=\cos{\left(x \right)}$$$:
$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\sec{\left(x \right)} d x} = \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)} + \int{\sec{\left(x \right)} d x}$$
Reescribe la secante como $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:
$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\sec{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}$$
Expresa el coseno en función del seno utilizando la fórmula $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ y luego expresa el seno utilizando la fórmula del ángulo doble $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:
$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}$$
Multiplica el numerador y el denominador por $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:
$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}$$
Sea $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$.
Entonces $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.
La integral se convierte en
$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{u} d u}}}$$
La integral de $$$\frac{1}{u}$$$ es $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{u} d u}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
Recordemos que $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:
$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}$$
Por lo tanto,
$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}$$
Añade la constante de integración:
$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)d x} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}+C$$
Respuesta
$$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)\, dx = \left(\ln\left(\left|{\cos{\left(x \right)}}\right|\right) + \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right)\right) + C$$$A