Funktion $$$e^{\sec^{2}{\left(x \right)}} \tan{\left(x \right)}$$$ integraali

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

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

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

Siis,

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

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

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

Tällä integraalilla (Eksponentti-integraali) ei ole suljettua muotoa:

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

Muista, että $$$u=\sec^{2}{\left(x \right)}$$$:

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

Näin ollen,

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

Lisää integrointivakio:

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

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