Integral von $$$\tan{\left(x \right)} \sec^{2}{\left(x \right)}$$$

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

Sei $$$u=\tan{\left(x \right)}$$$.

Dann $$$du=\left(\tan{\left(x \right)}\right)^{\prime }dx = \sec^{2}{\left(x \right)} dx$$$ (die Schritte sind » zu sehen), und es gilt $$$\sec^{2}{\left(x \right)} dx = du$$$.

Somit,

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

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=1$$$ an:

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

Zur Erinnerung: $$$u=\tan{\left(x \right)}$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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