Integral de $$$\sec^{5}{\left(x \right)}$$$

Encontre $$$\int \sec^{5}{\left(x \right)}\, dx$$$.

Para a integral $$$\int{\sec^{5}{\left(x \right)} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=\sec^{3}{\left(x \right)}$$$ e $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Então $$$\operatorname{du}=\left(\sec^{3}{\left(x \right)}\right)^{\prime }dx=3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (os passos podem ser vistos »).

Logo,

$$\int{\sec^{5}{\left(x \right)} d x}=\sec^{3}{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot 3 \tan{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - \int{3 \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}$$

Coloque a constante em evidência:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - \int{3 \tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}$$

Aplique a fórmula $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\tan^{2}{\left(x \right)} \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{3}{\left(x \right)} d x}$$

Expandir:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec^{3}{\left(x \right)} d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{5}{\left(x \right)} - \sec^{3}{\left(x \right)}\right)d x}$$

A integral de uma diferença é a diferença das integrais:

$$\tan{\left(x \right)} \sec^{3}{\left(x \right)} - 3 \int{\left(\sec^{5}{\left(x \right)} - \sec^{3}{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} + 3 \int{\sec^{3}{\left(x \right)} d x} - 3 \int{\sec^{5}{\left(x \right)} d x}$$

Assim, obtemos a seguinte equação linear simples em relação à integral:

$${\color{red}{\int{\sec^{5}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec^{3}{\left(x \right)} + 3 \int{\sec^{3}{\left(x \right)} d x} - 3 {\color{red}{\int{\sec^{5}{\left(x \right)} d x}}}$$

Resolvendo, obtemos que

$$\int{\sec^{5}{\left(x \right)} d x}=\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \int{\sec^{3}{\left(x \right)} d x}}{4}$$

Para a integral $$$\int{\sec^{3}{\left(x \right)} d x}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=\sec{\left(x \right)}$$$ e $$$\operatorname{dv}=\sec^{2}{\left(x \right)} dx$$$.

Então $$$\operatorname{du}=\left(\sec{\left(x \right)}\right)^{\prime }dx=\tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{\sec^{2}{\left(x \right)} d x}=\tan{\left(x \right)}$$$ (os passos podem ser vistos »).

Logo,

$$\int{\sec^{3}{\left(x \right)} d x}=\sec{\left(x \right)} \cdot \tan{\left(x \right)}-\int{\tan{\left(x \right)} \cdot \tan{\left(x \right)} \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\tan^{2}{\left(x \right)} \sec{\left(x \right)} d x}$$

Aplique a fórmula $$$\tan^{2}{\left(x \right)} = \sec^{2}{\left(x \right)} - 1$$$:

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

Expandir:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{2}{\left(x \right)} - 1\right) \sec{\left(x \right)} d x}=\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}$$

A integral de uma diferença é a diferença das integrais:

$$\tan{\left(x \right)} \sec{\left(x \right)} - \int{\left(\sec^{3}{\left(x \right)} - \sec{\left(x \right)}\right)d x}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - \int{\sec^{3}{\left(x \right)} d x}$$

Assim, obtemos a seguinte equação linear simples em relação à integral:

$${\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}=\tan{\left(x \right)} \sec{\left(x \right)} + \int{\sec{\left(x \right)} d x} - {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}$$

Resolvendo, obtemos que

$$\int{\sec^{3}{\left(x \right)} d x}=\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{\int{\sec{\left(x \right)} d x}}{2}$$

Portanto,

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 {\color{red}{\int{\sec^{3}{\left(x \right)} d x}}}}{4} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 {\color{red}{\left(\frac{\tan{\left(x \right)} \sec{\left(x \right)}}{2} + \frac{\int{\sec{\left(x \right)} d x}}{2}\right)}}}{4}$$

Reescreva a secante como $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\sec{\left(x \right)} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{8}$$

Reescreva o cosseno em termos do seno usando a fórmula $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ e depois reescreva o seno usando a fórmula do ângulo duplo $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8}$$

Multiplique o numerador e o denominador por $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8}$$

Seja $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$.

Então $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (veja os passos »), e obtemos $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

Portanto,

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

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

$$\frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\int{\frac{1}{u} d u}}}}{8} = \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} + \frac{3 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

Recorde que $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:

$$\frac{3 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8} = \frac{3 \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}$$

Portanto,

$$\int{\sec^{5}{\left(x \right)} d x} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}$$

Adicione a constante de integração:

$$\int{\sec^{5}{\left(x \right)} d x} = \frac{3 \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)}}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}+C$$

$$$\int \sec^{5}{\left(x \right)}\, dx = \left(\frac{3 \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right)}{8} + \frac{\tan{\left(x \right)} \sec^{3}{\left(x \right)}}{4} + \frac{3 \tan{\left(x \right)} \sec{\left(x \right)}}{8}\right) + C$$$A