Integralen av $$$\frac{1}{\sqrt{y \left(y - 1\right)}}$$$

Bestäm $$$\int \frac{1}{\sqrt{y \left(y - 1\right)}}\, dy$$$.

Inmatningen skrivs om: $$$\int{\frac{1}{\sqrt{y \left(y - 1\right)}} d y}=\int{\frac{1}{\sqrt{y^{2} - y}} d y}$$$.

Kvadratkomplettera (stegen kan ses »): $$$y^{2} - y = \left(y - \frac{1}{2}\right)^{2} - \frac{1}{4}$$$:

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

Låt $$$u=y - \frac{1}{2}$$$ vara.

Då $$$du=\left(y - \frac{1}{2}\right)^{\prime }dy = 1 dy$$$ (stegen kan ses »), och vi har att $$$dy = du$$$.

Alltså,

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

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

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

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

Alltså,

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

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

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

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

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

Alltså,

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

Tillämpa konstantregeln $$$\int c\, dv = c v$$$ med $$$c=1$$$:

$${\color{red}{\int{1 d v}}} = {\color{red}{v}}$$

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

$${\color{red}{v}} = {\color{red}{\operatorname{acosh}{\left(2 u \right)}}}$$

Kom ihåg att $$$u=y - \frac{1}{2}$$$:

$$\operatorname{acosh}{\left(2 {\color{red}{u}} \right)} = \operatorname{acosh}{\left(2 {\color{red}{\left(y - \frac{1}{2}\right)}} \right)}$$

Alltså,

$$\int{\frac{1}{\sqrt{y^{2} - y}} d y} = \operatorname{acosh}{\left(2 y - 1 \right)}$$

Lägg till integrationskonstanten:

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

$$$\int \frac{1}{\sqrt{y \left(y - 1\right)}}\, dy = \operatorname{acosh}{\left(2 y - 1 \right)} + C$$$A