Integral de $$$\frac{4}{\sqrt{t^{2} - 4}}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \frac{4}{\sqrt{t^{2} - 4}}\, dt$$$.
Solución
Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=4$$$ y $$$f{\left(t \right)} = \frac{1}{\sqrt{t^{2} - 4}}$$$:
$${\color{red}{\int{\frac{4}{\sqrt{t^{2} - 4}} d t}}} = {\color{red}{\left(4 \int{\frac{1}{\sqrt{t^{2} - 4}} d t}\right)}}$$
Sea $$$t=2 \cosh{\left(u \right)}$$$.
Entonces $$$dt=\left(2 \cosh{\left(u \right)}\right)^{\prime }du = 2 \sinh{\left(u \right)} du$$$ (los pasos pueden verse »).
Además, se sigue que $$$u=\operatorname{acosh}{\left(\frac{t}{2} \right)}$$$.
El integrando se convierte en
$$$\frac{1}{\sqrt{t^{2} - 4}} = \frac{1}{\sqrt{4 \cosh^{2}{\left( u \right)} - 4}}$$$
Utiliza la identidad $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:
$$$\frac{1}{\sqrt{4 \cosh^{2}{\left( u \right)} - 4}}=\frac{1}{2 \sqrt{\cosh^{2}{\left( u \right)} - 1}}=\frac{1}{2 \sqrt{\sinh^{2}{\left( u \right)}}}$$$
Suponiendo que $$$\sinh{\left( u \right)} \ge 0$$$, obtenemos lo siguiente:
$$$\frac{1}{2 \sqrt{\sinh^{2}{\left( u \right)}}} = \frac{1}{2 \sinh{\left( u \right)}}$$$
Por lo tanto,
$$4 {\color{red}{\int{\frac{1}{\sqrt{t^{2} - 4}} d t}}} = 4 {\color{red}{\int{1 d u}}}$$
Aplica la regla de la constante $$$\int c\, du = c u$$$ con $$$c=1$$$:
$$4 {\color{red}{\int{1 d u}}} = 4 {\color{red}{u}}$$
Recordemos que $$$u=\operatorname{acosh}{\left(\frac{t}{2} \right)}$$$:
$$4 {\color{red}{u}} = 4 {\color{red}{\operatorname{acosh}{\left(\frac{t}{2} \right)}}}$$
Por lo tanto,
$$\int{\frac{4}{\sqrt{t^{2} - 4}} d t} = 4 \operatorname{acosh}{\left(\frac{t}{2} \right)}$$
Añade la constante de integración:
$$\int{\frac{4}{\sqrt{t^{2} - 4}} d t} = 4 \operatorname{acosh}{\left(\frac{t}{2} \right)}+C$$
Respuesta
$$$\int \frac{4}{\sqrt{t^{2} - 4}}\, dt = 4 \operatorname{acosh}{\left(\frac{t}{2} \right)} + C$$$A