Integral von $$$\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)}$$$
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Ihre Eingabe
Bestimme $$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)}\, dx$$$.
Lösung
Expand the expression:
$${\color{red}{\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)} d x}}} = {\color{red}{\int{\left(- \tan{\left(x \right)} \sec{\left(x \right)} + \sec^{2}{\left(x \right)}\right)d x}}}$$
Gliedweise integrieren:
$${\color{red}{\int{\left(- \tan{\left(x \right)} \sec{\left(x \right)} + \sec^{2}{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{\tan{\left(x \right)} \sec{\left(x \right)} d x} + \int{\sec^{2}{\left(x \right)} d x}\right)}}$$
Das Integral von $$$\sec^{2}{\left(x \right)}$$$ ist $$$\int{\sec^{2}{\left(x \right)} d x} = \tan{\left(x \right)}$$$:
$$- \int{\tan{\left(x \right)} \sec{\left(x \right)} d x} + {\color{red}{\int{\sec^{2}{\left(x \right)} d x}}} = - \int{\tan{\left(x \right)} \sec{\left(x \right)} d x} + {\color{red}{\tan{\left(x \right)}}}$$
Das Integral von $$$\tan{\left(x \right)} \sec{\left(x \right)}$$$ ist $$$\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} = \sec{\left(x \right)}$$$:
$$\tan{\left(x \right)} - {\color{red}{\int{\tan{\left(x \right)} \sec{\left(x \right)} d x}}} = \tan{\left(x \right)} - {\color{red}{\sec{\left(x \right)}}}$$
Daher,
$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)} d x} = \tan{\left(x \right)} - \sec{\left(x \right)}$$
Fügen Sie die Integrationskonstante hinzu:
$$\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)} d x} = \tan{\left(x \right)} - \sec{\left(x \right)}+C$$
Antwort
$$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)}\, dx = \left(\tan{\left(x \right)} - \sec{\left(x \right)}\right) + C$$$A