Integralen av $$$\frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}}$$$

Bestäm $$$\int \frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}}\, dx$$$.

Skriv om integranden i termer av sinus och/eller cosinus:

$${\color{red}{\int{\frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}} d x}}}$$

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

Då $$$du=\left(\sin{\left(x \right)}\right)^{\prime }dx = \cos{\left(x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\cos{\left(x \right)} dx = du$$$.

Alltså,

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-2$$$:

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

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

$$- {\color{red}{u}}^{-1} = - {\color{red}{\sin{\left(x \right)}}}^{-1}$$

Alltså,

$$\int{\frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x} = - \frac{1}{\sin{\left(x \right)}}$$

Lägg till integrationskonstanten:

$$\int{\frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}} d x} = - \frac{1}{\sin{\left(x \right)}}+C$$

$$$\int \frac{\sec{\left(x \right)}}{\tan^{2}{\left(x \right)}}\, dx = - \frac{1}{\sin{\left(x \right)}} + C$$$A