Integral von $$$- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}$$$

Bestimme $$$\int \left(- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}\right)\, dx$$$.

Gliedweise integrieren:

$${\color{red}{\int{\left(- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}\right)d x}}} = {\color{red}{\left(- \int{\csc^{2}{\left(6 x \right)} d x} + \int{\sec^{2}{\left(5 x \right)} d x}\right)}}$$

Sei $$$u=5 x$$$.

Dann $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{5}$$$.

Das Integral wird zu

$$- \int{\csc^{2}{\left(6 x \right)} d x} + {\color{red}{\int{\sec^{2}{\left(5 x \right)} d x}}} = - \int{\csc^{2}{\left(6 x \right)} d x} + {\color{red}{\int{\frac{\sec^{2}{\left(u \right)}}{5} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{5}$$$ und $$$f{\left(u \right)} = \sec^{2}{\left(u \right)}$$$ an:

$$- \int{\csc^{2}{\left(6 x \right)} d x} + {\color{red}{\int{\frac{\sec^{2}{\left(u \right)}}{5} d u}}} = - \int{\csc^{2}{\left(6 x \right)} d x} + {\color{red}{\left(\frac{\int{\sec^{2}{\left(u \right)} d u}}{5}\right)}}$$

Das Integral von $$$\sec^{2}{\left(u \right)}$$$ ist $$$\int{\sec^{2}{\left(u \right)} d u} = \tan{\left(u \right)}$$$:

$$- \int{\csc^{2}{\left(6 x \right)} d x} + \frac{{\color{red}{\int{\sec^{2}{\left(u \right)} d u}}}}{5} = - \int{\csc^{2}{\left(6 x \right)} d x} + \frac{{\color{red}{\tan{\left(u \right)}}}}{5}$$

Zur Erinnerung: $$$u=5 x$$$:

$$- \int{\csc^{2}{\left(6 x \right)} d x} + \frac{\tan{\left({\color{red}{u}} \right)}}{5} = - \int{\csc^{2}{\left(6 x \right)} d x} + \frac{\tan{\left({\color{red}{\left(5 x\right)}} \right)}}{5}$$

Sei $$$u=6 x$$$.

Dann $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = \frac{du}{6}$$$.

Also,

$$\frac{\tan{\left(5 x \right)}}{5} - {\color{red}{\int{\csc^{2}{\left(6 x \right)} d x}}} = \frac{\tan{\left(5 x \right)}}{5} - {\color{red}{\int{\frac{\csc^{2}{\left(u \right)}}{6} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{6}$$$ und $$$f{\left(u \right)} = \csc^{2}{\left(u \right)}$$$ an:

$$\frac{\tan{\left(5 x \right)}}{5} - {\color{red}{\int{\frac{\csc^{2}{\left(u \right)}}{6} d u}}} = \frac{\tan{\left(5 x \right)}}{5} - {\color{red}{\left(\frac{\int{\csc^{2}{\left(u \right)} d u}}{6}\right)}}$$

Das Integral von $$$\csc^{2}{\left(u \right)}$$$ ist $$$\int{\csc^{2}{\left(u \right)} d u} = - \cot{\left(u \right)}$$$:

$$\frac{\tan{\left(5 x \right)}}{5} - \frac{{\color{red}{\int{\csc^{2}{\left(u \right)} d u}}}}{6} = \frac{\tan{\left(5 x \right)}}{5} - \frac{{\color{red}{\left(- \cot{\left(u \right)}\right)}}}{6}$$

Zur Erinnerung: $$$u=6 x$$$:

$$\frac{\tan{\left(5 x \right)}}{5} + \frac{\cot{\left({\color{red}{u}} \right)}}{6} = \frac{\tan{\left(5 x \right)}}{5} + \frac{\cot{\left({\color{red}{\left(6 x\right)}} \right)}}{6}$$

Daher,

$$\int{\left(- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}\right)d x} = \frac{\tan{\left(5 x \right)}}{5} + \frac{\cot{\left(6 x \right)}}{6}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}\right)d x} = \frac{\tan{\left(5 x \right)}}{5} + \frac{\cot{\left(6 x \right)}}{6}+C$$

$$$\int \left(- \csc^{2}{\left(6 x \right)} + \sec^{2}{\left(5 x \right)}\right)\, dx = \left(\frac{\tan{\left(5 x \right)}}{5} + \frac{\cot{\left(6 x \right)}}{6}\right) + C$$$A