Integral de $$$\sec{\left(4 x \right)}$$$
Calculadora relacionada: Calculadora de Integrais Definidas e Impróprias
Sua entrada
Encontre $$$\int \sec{\left(4 x \right)}\, dx$$$.
Solução
Seja $$$u=4 x$$$.
Então $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{4}$$$.
Portanto,
$${\color{red}{\int{\sec{\left(4 x \right)} d x}}} = {\color{red}{\int{\frac{\sec{\left(u \right)}}{4} d u}}}$$
Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{4}$$$ e $$$f{\left(u \right)} = \sec{\left(u \right)}$$$:
$${\color{red}{\int{\frac{\sec{\left(u \right)}}{4} d u}}} = {\color{red}{\left(\frac{\int{\sec{\left(u \right)} d u}}{4}\right)}}$$
Reescreva a secante como $$$\sec\left( u \right)=\frac{1}{\cos\left( u \right)}$$$:
$$\frac{{\color{red}{\int{\sec{\left(u \right)} d u}}}}{4} = \frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{4}$$
Reescreva o cosseno em termos do seno usando a fórmula $$$\cos\left( u \right)=\sin\left( u + \frac{\pi}{2}\right)$$$ e depois reescreva o seno usando a fórmula do ângulo duplo $$$\sin\left( u \right)=2\sin\left(\frac{ u }{2}\right)\cos\left(\frac{ u }{2}\right)$$$:
$$\frac{{\color{red}{\int{\frac{1}{\cos{\left(u \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{4}$$
Multiplique o numerador e o denominador por $$$\sec^2\left(\frac{ u }{2} + \frac{\pi}{4} \right)$$$:
$$\frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{u}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{4}$$
Seja $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$.
Então $$$dv=\left(\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}\right)^{\prime }du = \frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2} du$$$ (veja os passos »), e obtemos $$$\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)} du = 2 dv$$$.
Logo,
$$\frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4}$$
A integral de $$$\frac{1}{v}$$$ é $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:
$$\frac{{\color{red}{\int{\frac{1}{v} d v}}}}{4} = \frac{{\color{red}{\ln{\left(\left|{v}\right| \right)}}}}{4}$$
Recorde que $$$v=\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}$$$:
$$\frac{\ln{\left(\left|{{\color{red}{v}}}\right| \right)}}{4} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{u}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{4}$$
Recorde que $$$u=4 x$$$:
$$\frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{u}}}{2} \right)}}\right| \right)}}{4} = \frac{\ln{\left(\left|{\tan{\left(\frac{\pi}{4} + \frac{{\color{red}{\left(4 x\right)}}}{2} \right)}}\right| \right)}}{4}$$
Portanto,
$$\int{\sec{\left(4 x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(2 x + \frac{\pi}{4} \right)}}\right| \right)}}{4}$$
Adicione a constante de integração:
$$\int{\sec{\left(4 x \right)} d x} = \frac{\ln{\left(\left|{\tan{\left(2 x + \frac{\pi}{4} \right)}}\right| \right)}}{4}+C$$
Resposta
$$$\int \sec{\left(4 x \right)}\, dx = \frac{\ln\left(\left|{\tan{\left(2 x + \frac{\pi}{4} \right)}}\right|\right)}{4} + C$$$A