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

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

Låt $$$u=\sec^{2}{\left(x \right)}$$$ vara.

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

Alltså,

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

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

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

Denna integral (Exponentialintegralen) har ingen sluten form:

$$\frac{{\color{red}{\int{\frac{e^{u}}{u} d u}}}}{2} = \frac{{\color{red}{\operatorname{Ei}{\left(u \right)}}}}{2}$$

Kom ihåg att $$$u=\sec^{2}{\left(x \right)}$$$:

$$\frac{\operatorname{Ei}{\left({\color{red}{u}} \right)}}{2} = \frac{\operatorname{Ei}{\left({\color{red}{\sec^{2}{\left(x \right)}}} \right)}}{2}$$

Alltså,

$$\int{e^{\sec^{2}{\left(x \right)}} \tan{\left(x \right)} d x} = \frac{\operatorname{Ei}{\left(\sec^{2}{\left(x \right)} \right)}}{2}$$

Lägg till integrationskonstanten:

$$\int{e^{\sec^{2}{\left(x \right)}} \tan{\left(x \right)} d x} = \frac{\operatorname{Ei}{\left(\sec^{2}{\left(x \right)} \right)}}{2}+C$$

$$$\int e^{\sec^{2}{\left(x \right)}} \tan{\left(x \right)}\, dx = \frac{\operatorname{Ei}{\left(\sec^{2}{\left(x \right)} \right)}}{2} + C$$$A