Integral de $$$\sqrt{9 x^{2} - 1}$$$

Encontre $$$\int \sqrt{9 x^{2} - 1}\, dx$$$.

Seja $$$x=\frac{\cosh{\left(u \right)}}{3}$$$.

Então $$$dx=\left(\frac{\cosh{\left(u \right)}}{3}\right)^{\prime }du = \frac{\sinh{\left(u \right)}}{3} du$$$ (os passos podem ser vistos »).

Além disso, segue-se que $$$u=\operatorname{acosh}{\left(3 x \right)}$$$.

Portanto,

$$$\sqrt{9 x^{2} - 1} = \sqrt{\cosh^{2}{\left( u \right)} - 1}$$$

Use a identidade $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

$$$\sqrt{\cosh^{2}{\left( u \right)} - 1}=\sqrt{\sinh^{2}{\left( u \right)}}$$$

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

$$$\sqrt{\sinh^{2}{\left( u \right)}} = \sinh{\left( u \right)}$$$

Portanto,

$${\color{red}{\int{\sqrt{9 x^{2} - 1} d x}}} = {\color{red}{\int{\frac{\sinh^{2}{\left(u \right)}}{3} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{3}$$$ e $$$f{\left(u \right)} = \sinh^{2}{\left(u \right)}$$$:

$${\color{red}{\int{\frac{\sinh^{2}{\left(u \right)}}{3} d u}}} = {\color{red}{\left(\frac{\int{\sinh^{2}{\left(u \right)} d u}}{3}\right)}}$$

Aplique a fórmula de redução de potência $$$\sinh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} - \frac{1}{2}$$$ com $$$\alpha= u $$$:

$$\frac{{\color{red}{\int{\sinh^{2}{\left(u \right)} d u}}}}{3} = \frac{{\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}}}{3}$$

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

$$\frac{{\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}}}{3} = \frac{{\color{red}{\left(\frac{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}{2}\right)}}}{3}$$

Integre termo a termo:

$$\frac{{\color{red}{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}}}{6} = \frac{{\color{red}{\left(- \int{1 d u} + \int{\cosh{\left(2 u \right)} d u}\right)}}}{6}$$

Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:

$$\frac{\int{\cosh{\left(2 u \right)} d u}}{6} - \frac{{\color{red}{\int{1 d u}}}}{6} = \frac{\int{\cosh{\left(2 u \right)} d u}}{6} - \frac{{\color{red}{u}}}{6}$$

Seja $$$v=2 u$$$.

Então $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (veja os passos »), e obtemos $$$du = \frac{dv}{2}$$$.

A integral pode ser reescrita como

$$- \frac{u}{6} + \frac{{\color{red}{\int{\cosh{\left(2 u \right)} d u}}}}{6} = - \frac{u}{6} + \frac{{\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{6}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(v \right)} = \cosh{\left(v \right)}$$$:

$$- \frac{u}{6} + \frac{{\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{6} = - \frac{u}{6} + \frac{{\color{red}{\left(\frac{\int{\cosh{\left(v \right)} d v}}{2}\right)}}}{6}$$

A integral do cosseno hiperbólico é $$$\int{\cosh{\left(v \right)} d v} = \sinh{\left(v \right)}$$$:

$$- \frac{u}{6} + \frac{{\color{red}{\int{\cosh{\left(v \right)} d v}}}}{12} = - \frac{u}{6} + \frac{{\color{red}{\sinh{\left(v \right)}}}}{12}$$

Recorde que $$$v=2 u$$$:

$$- \frac{u}{6} + \frac{\sinh{\left({\color{red}{v}} \right)}}{12} = - \frac{u}{6} + \frac{\sinh{\left({\color{red}{\left(2 u\right)}} \right)}}{12}$$

Recorde que $$$u=\operatorname{acosh}{\left(3 x \right)}$$$:

$$\frac{\sinh{\left(2 {\color{red}{u}} \right)}}{12} - \frac{{\color{red}{u}}}{6} = \frac{\sinh{\left(2 {\color{red}{\operatorname{acosh}{\left(3 x \right)}}} \right)}}{12} - \frac{{\color{red}{\operatorname{acosh}{\left(3 x \right)}}}}{6}$$

Portanto,

$$\int{\sqrt{9 x^{2} - 1} d x} = \frac{\sinh{\left(2 \operatorname{acosh}{\left(3 x \right)} \right)}}{12} - \frac{\operatorname{acosh}{\left(3 x \right)}}{6}$$

Usando as fórmulas $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, simplifique a expressão:

$$\int{\sqrt{9 x^{2} - 1} d x} = \frac{x \sqrt{3 x - 1} \sqrt{3 x + 1}}{2} - \frac{\operatorname{acosh}{\left(3 x \right)}}{6}$$

Adicione a constante de integração:

$$\int{\sqrt{9 x^{2} - 1} d x} = \frac{x \sqrt{3 x - 1} \sqrt{3 x + 1}}{2} - \frac{\operatorname{acosh}{\left(3 x \right)}}{6}+C$$

$$$\int \sqrt{9 x^{2} - 1}\, dx = \left(\frac{x \sqrt{3 x - 1} \sqrt{3 x + 1}}{2} - \frac{\operatorname{acosh}{\left(3 x \right)}}{6}\right) + C$$$A