Integralen av $$$\frac{4}{\sqrt{t^{2} - 4}}$$$

Bestäm $$$\int \frac{4}{\sqrt{t^{2} - 4}}\, dt$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ med $$$c=4$$$ och $$$f{\left(t \right)} = \frac{1}{\sqrt{t^{2} - 4}}$$$:

$${\color{red}{\int{\frac{4}{\sqrt{t^{2} - 4}} d t}}} = {\color{red}{\left(4 \int{\frac{1}{\sqrt{t^{2} - 4}} d t}\right)}}$$

Låt $$$t=2 \cosh{\left(u \right)}$$$ vara.

Då $$$dt=\left(2 \cosh{\left(u \right)}\right)^{\prime }du = 2 \sinh{\left(u \right)} du$$$ (stegen kan ses »).

Det följer också att $$$u=\operatorname{acosh}{\left(\frac{t}{2} \right)}$$$.

Alltså,

$$$\frac{1}{\sqrt{t^{2} - 4}} = \frac{1}{\sqrt{4 \cosh^{2}{\left( u \right)} - 4}}$$$

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

$$$\frac{1}{\sqrt{4 \cosh^{2}{\left( u \right)} - 4}}=\frac{1}{2 \sqrt{\cosh^{2}{\left( u \right)} - 1}}=\frac{1}{2 \sqrt{\sinh^{2}{\left( u \right)}}}$$$

Om vi antar att $$$\sinh{\left( u \right)} \ge 0$$$, erhåller vi följande:

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

Integralen blir

$$4 {\color{red}{\int{\frac{1}{\sqrt{t^{2} - 4}} d t}}} = 4 {\color{red}{\int{1 d u}}}$$

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=1$$$:

$$4 {\color{red}{\int{1 d u}}} = 4 {\color{red}{u}}$$

Kom ihåg att $$$u=\operatorname{acosh}{\left(\frac{t}{2} \right)}$$$:

$$4 {\color{red}{u}} = 4 {\color{red}{\operatorname{acosh}{\left(\frac{t}{2} \right)}}}$$

Alltså,

$$\int{\frac{4}{\sqrt{t^{2} - 4}} d t} = 4 \operatorname{acosh}{\left(\frac{t}{2} \right)}$$

Lägg till integrationskonstanten:

$$\int{\frac{4}{\sqrt{t^{2} - 4}} d t} = 4 \operatorname{acosh}{\left(\frac{t}{2} \right)}+C$$

$$$\int \frac{4}{\sqrt{t^{2} - 4}}\, dt = 4 \operatorname{acosh}{\left(\frac{t}{2} \right)} + C$$$A