Integralen av $$$\sqrt{x^{2} - 25}$$$

Bestäm $$$\int \sqrt{x^{2} - 25}\, dx$$$.

Låt $$$x=5 \cosh{\left(u \right)}$$$ vara.

Då $$$dx=\left(5 \cosh{\left(u \right)}\right)^{\prime }du = 5 \sinh{\left(u \right)} du$$$ (stegen kan ses »).

Det följer också att $$$u=\operatorname{acosh}{\left(\frac{x}{5} \right)}$$$.

Alltså,

$$$\sqrt{x^{2} - 25} = \sqrt{25 \cosh^{2}{\left( u \right)} - 25}$$$

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

$$$\sqrt{25 \cosh^{2}{\left( u \right)} - 25}=5 \sqrt{\cosh^{2}{\left( u \right)} - 1}=5 \sqrt{\sinh^{2}{\left( u \right)}}$$$

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

$$$5 \sqrt{\sinh^{2}{\left( u \right)}} = 5 \sinh{\left( u \right)}$$$

Alltså,

$${\color{red}{\int{\sqrt{x^{2} - 25} d x}}} = {\color{red}{\int{25 \sinh^{2}{\left(u \right)} d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=25$$$ och $$$f{\left(u \right)} = \sinh^{2}{\left(u \right)}$$$:

$${\color{red}{\int{25 \sinh^{2}{\left(u \right)} d u}}} = {\color{red}{\left(25 \int{\sinh^{2}{\left(u \right)} d u}\right)}}$$

Använd potensreduceringsformeln $$$\sinh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} - \frac{1}{2}$$$ med $$$\alpha= u $$$:

$$25 {\color{red}{\int{\sinh^{2}{\left(u \right)} d u}}} = 25 {\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = \cosh{\left(2 u \right)} - 1$$$:

$$25 {\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}} = 25 {\color{red}{\left(\frac{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}{2}\right)}}$$

Integrera termvis:

$$\frac{25 {\color{red}{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}}}{2} = \frac{25 {\color{red}{\left(- \int{1 d u} + \int{\cosh{\left(2 u \right)} d u}\right)}}}{2}$$

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

$$\frac{25 \int{\cosh{\left(2 u \right)} d u}}{2} - \frac{25 {\color{red}{\int{1 d u}}}}{2} = \frac{25 \int{\cosh{\left(2 u \right)} d u}}{2} - \frac{25 {\color{red}{u}}}{2}$$

Låt $$$v=2 u$$$ vara.

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

Alltså,

$$- \frac{25 u}{2} + \frac{25 {\color{red}{\int{\cosh{\left(2 u \right)} d u}}}}{2} = - \frac{25 u}{2} + \frac{25 {\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{2}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(v \right)} = \cosh{\left(v \right)}$$$:

$$- \frac{25 u}{2} + \frac{25 {\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{2} = - \frac{25 u}{2} + \frac{25 {\color{red}{\left(\frac{\int{\cosh{\left(v \right)} d v}}{2}\right)}}}{2}$$

Integralen av den hyperboliska cosinusfunktionen är $$$\int{\cosh{\left(v \right)} d v} = \sinh{\left(v \right)}$$$:

$$- \frac{25 u}{2} + \frac{25 {\color{red}{\int{\cosh{\left(v \right)} d v}}}}{4} = - \frac{25 u}{2} + \frac{25 {\color{red}{\sinh{\left(v \right)}}}}{4}$$

Kom ihåg att $$$v=2 u$$$:

$$- \frac{25 u}{2} + \frac{25 \sinh{\left({\color{red}{v}} \right)}}{4} = - \frac{25 u}{2} + \frac{25 \sinh{\left({\color{red}{\left(2 u\right)}} \right)}}{4}$$

Kom ihåg att $$$u=\operatorname{acosh}{\left(\frac{x}{5} \right)}$$$:

$$\frac{25 \sinh{\left(2 {\color{red}{u}} \right)}}{4} - \frac{25 {\color{red}{u}}}{2} = \frac{25 \sinh{\left(2 {\color{red}{\operatorname{acosh}{\left(\frac{x}{5} \right)}}} \right)}}{4} - \frac{25 {\color{red}{\operatorname{acosh}{\left(\frac{x}{5} \right)}}}}{2}$$

Alltså,

$$\int{\sqrt{x^{2} - 25} d x} = \frac{25 \sinh{\left(2 \operatorname{acosh}{\left(\frac{x}{5} \right)} \right)}}{4} - \frac{25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2}$$

Använd formlerna $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$ för att förenkla uttrycket:

$$\int{\sqrt{x^{2} - 25} d x} = \frac{5 x \sqrt{\frac{x}{5} - 1} \sqrt{\frac{x}{5} + 1}}{2} - \frac{25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2}$$

Förenkla ytterligare:

$$\int{\sqrt{x^{2} - 25} d x} = \frac{x \sqrt{x - 5} \sqrt{x + 5} - 25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2}$$

Lägg till integrationskonstanten:

$$\int{\sqrt{x^{2} - 25} d x} = \frac{x \sqrt{x - 5} \sqrt{x + 5} - 25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2}+C$$

$$$\int \sqrt{x^{2} - 25}\, dx = \frac{x \sqrt{x - 5} \sqrt{x + 5} - 25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2} + C$$$A