Integralen av $$$\frac{1}{x^{2} \sqrt{1 - x^{2}}}$$$

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

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

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

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

Alltså,

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

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

$$$\frac{1}{\sqrt{1 - \sin^{2}{\left( u \right)}} \sin^{2}{\left( u \right)}}=\frac{1}{\sqrt{\cos^{2}{\left( u \right)}} \sin^{2}{\left( u \right)}}$$$

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

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

Alltså,

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

Skriv om integranden i termer av kosekanten:

$${\color{red}{\int{\frac{1}{\sin^{2}{\left(u \right)}} d u}}} = {\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}$$

Integralen av $$$\csc^{2}{\left(u \right)}$$$ är $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$:

$${\color{red}{\int{\csc^{2}{\left(u \right)} d u}}} = {\color{red}{\left(- \cot{\left(u \right)}\right)}}$$

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

$$- \cot{\left({\color{red}{u}} \right)} = - \cot{\left({\color{red}{\operatorname{asin}{\left(x \right)}}} \right)}$$

Alltså,

$$\int{\frac{1}{x^{2} \sqrt{1 - x^{2}}} d x} = - \frac{\sqrt{1 - x^{2}}}{x}$$

Lägg till integrationskonstanten:

$$\int{\frac{1}{x^{2} \sqrt{1 - x^{2}}} d x} = - \frac{\sqrt{1 - x^{2}}}{x}+C$$

$$$\int \frac{1}{x^{2} \sqrt{1 - x^{2}}}\, dx = - \frac{\sqrt{1 - x^{2}}}{x} + C$$$A