Integral de $$$\frac{1}{\sqrt{- 2 t^{26} + t^{2}}}$$$

Encontre $$$\int \frac{1}{\sqrt{- 2 t^{26} + t^{2}}}\, dt$$$.

A entrada é reescrita como: $$$\int{\frac{1}{\sqrt{- 2 t^{26} + t^{2}}} d t}=\int{\frac{1}{t \sqrt{1 - 2 t^{24}}} d t}$$$.

Seja $$$u=t^{12}$$$.

Então $$$du=\left(t^{12}\right)^{\prime }dt = 12 t^{11} dt$$$ (veja os passos »), e obtemos $$$t^{11} dt = \frac{du}{12}$$$.

Assim,

$${\color{red}{\int{\frac{1}{t \sqrt{1 - 2 t^{24}}} d t}}} = {\color{red}{\int{\frac{1}{12 u \sqrt{1 - 2 u^{2}}} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{12}$$$ e $$$f{\left(u \right)} = \frac{1}{u \sqrt{1 - 2 u^{2}}}$$$:

$${\color{red}{\int{\frac{1}{12 u \sqrt{1 - 2 u^{2}}} d u}}} = {\color{red}{\left(\frac{\int{\frac{1}{u \sqrt{1 - 2 u^{2}}} d u}}{12}\right)}}$$

Seja $$$u=\frac{\sqrt{2} \sin{\left(v \right)}}{2}$$$.

Então $$$du=\left(\frac{\sqrt{2} \sin{\left(v \right)}}{2}\right)^{\prime }dv = \frac{\sqrt{2} \cos{\left(v \right)}}{2} dv$$$ (os passos podem ser vistos »).

Além disso, segue-se que $$$v=\operatorname{asin}{\left(\sqrt{2} u \right)}$$$.

O integrando torna-se

$$$\frac{1}{ u \sqrt{1 - 2 u ^{2}}} = \frac{\sqrt{2}}{\sqrt{1 - \sin^{2}{\left( v \right)}} \sin{\left( v \right)}}$$$

Use a identidade $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$:

$$$\frac{\sqrt{2}}{\sqrt{1 - \sin^{2}{\left( v \right)}} \sin{\left( v \right)}}=\frac{\sqrt{2}}{\sqrt{\cos^{2}{\left( v \right)}} \sin{\left( v \right)}}$$$

Supondo que $$$\cos{\left( v \right)} \ge 0$$$, obtemos o seguinte:

$$$\frac{\sqrt{2}}{\sqrt{\cos^{2}{\left( v \right)}} \sin{\left( v \right)}} = \frac{\sqrt{2}}{\sin{\left( v \right)} \cos{\left( v \right)}}$$$

Assim,

$$\frac{{\color{red}{\int{\frac{1}{u \sqrt{1 - 2 u^{2}}} d u}}}}{12} = \frac{{\color{red}{\int{\frac{1}{\sin{\left(v \right)}} d v}}}}{12}$$

Reescreva o seno usando a fórmula do ângulo duplo $$$\sin\left( v \right)=2\sin\left(\frac{ v }{2}\right)\cos\left(\frac{ v }{2}\right)$$$:

$$\frac{{\color{red}{\int{\frac{1}{\sin{\left(v \right)}} d v}}}}{12} = \frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{v}{2} \right)} \cos{\left(\frac{v}{2} \right)}} d v}}}}{12}$$

Multiplique o numerador e o denominador por $$$\sec^2\left(\frac{ v }{2} \right)$$$:

$$\frac{{\color{red}{\int{\frac{1}{2 \sin{\left(\frac{v}{2} \right)} \cos{\left(\frac{v}{2} \right)}} d v}}}}{12} = \frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{v}{2} \right)}}{2 \tan{\left(\frac{v}{2} \right)}} d v}}}}{12}$$

Seja $$$w=\tan{\left(\frac{v}{2} \right)}$$$.

Então $$$dw=\left(\tan{\left(\frac{v}{2} \right)}\right)^{\prime }dv = \frac{\sec^{2}{\left(\frac{v}{2} \right)}}{2} dv$$$ (veja os passos »), e obtemos $$$\sec^{2}{\left(\frac{v}{2} \right)} dv = 2 dw$$$.

A integral torna-se

$$\frac{{\color{red}{\int{\frac{\sec^{2}{\left(\frac{v}{2} \right)}}{2 \tan{\left(\frac{v}{2} \right)}} d v}}}}{12} = \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{12}$$

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

$$\frac{{\color{red}{\int{\frac{1}{w} d w}}}}{12} = \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{12}$$

Recorde que $$$w=\tan{\left(\frac{v}{2} \right)}$$$:

$$\frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{12} = \frac{\ln{\left(\left|{{\color{red}{\tan{\left(\frac{v}{2} \right)}}}}\right| \right)}}{12}$$

Recorde que $$$v=\operatorname{asin}{\left(\sqrt{2} u \right)}$$$:

$$\frac{\ln{\left(\left|{\tan{\left(\frac{{\color{red}{v}}}{2} \right)}}\right| \right)}}{12} = \frac{\ln{\left(\left|{\tan{\left(\frac{{\color{red}{\operatorname{asin}{\left(\sqrt{2} u \right)}}}}{2} \right)}}\right| \right)}}{12}$$

Recorde que $$$u=t^{12}$$$:

$$\frac{\ln{\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{2} {\color{red}{u}} \right)}}{2} \right)}}\right| \right)}}{12} = \frac{\ln{\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{2} {\color{red}{t^{12}}} \right)}}{2} \right)}}\right| \right)}}{12}$$

Portanto,

$$\int{\frac{1}{t \sqrt{1 - 2 t^{24}}} d t} = \frac{\ln{\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{2} t^{12} \right)}}{2} \right)}}\right| \right)}}{12}$$

Adicione a constante de integração:

$$\int{\frac{1}{t \sqrt{1 - 2 t^{24}}} d t} = \frac{\ln{\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{2} t^{12} \right)}}{2} \right)}}\right| \right)}}{12}+C$$

$$$\int \frac{1}{\sqrt{- 2 t^{26} + t^{2}}}\, dt = \frac{\ln\left(\left|{\tan{\left(\frac{\operatorname{asin}{\left(\sqrt{2} t^{12} \right)}}{2} \right)}}\right|\right)}{12} + C$$$A