Funktion $$$\sqrt{x^{2} - 1}$$$ integraali

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

Olkoon $$$x=\cosh{\left(u \right)}$$$.

Tällöin $$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$ (ratkaisuvaiheet ovat nähtävissä »).

Lisäksi seuraa, että $$$u=\operatorname{acosh}{\left(x \right)}$$$.

Näin ollen,

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

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

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

Olettamalla, että $$$\sinh{\left( u \right)} \ge 0$$$, saamme seuraavaa:

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

Siis,

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

Sovella potenssin alentamiskaavaa $$$\sinh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} - \frac{1}{2}$$$ käyttäen $$$\alpha= u $$$:

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(u \right)} = \cosh{\left(2 u \right)} - 1$$$:

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

Integroi termi kerrallaan:

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

Sovella vakiosääntöä $$$\int c\, du = c u$$$ käyttäen $$$c=1$$$:

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

Olkoon $$$v=2 u$$$.

Tällöin $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$du = \frac{dv}{2}$$$.

Integraali muuttuu muotoon

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(v \right)} = \cosh{\left(v \right)}$$$:

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

Hyperbolisen kosinin integraali on $$$\int{\cosh{\left(v \right)} d v} = \sinh{\left(v \right)}$$$:

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

Muista, että $$$v=2 u$$$:

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

Muista, että $$$u=\operatorname{acosh}{\left(x \right)}$$$:

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

Näin ollen,

$$\int{\sqrt{x^{2} - 1} d x} = \frac{\sinh{\left(2 \operatorname{acosh}{\left(x \right)} \right)}}{4} - \frac{\operatorname{acosh}{\left(x \right)}}{2}$$

Käyttämällä kaavoja $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, sievennä lauseke:

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

Sievennä edelleen:

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

Lisää integrointivakio:

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

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