Integral von $$$\frac{\sin{\left(x \right)}}{x + 1}$$$

Bestimme $$$\int \frac{\sin{\left(x \right)}}{x + 1}\, dx$$$.

Sei $$$u=x + 1$$$.

Dann $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = du$$$.

Das Integral lässt sich umschreiben als

$${\color{red}{\int{\frac{\sin{\left(x \right)}}{x + 1} d x}}} = {\color{red}{\int{\frac{\sin{\left(u - 1 \right)}}{u} d u}}}$$

Schreiben Sie den Integranden um:

$${\color{red}{\int{\frac{\sin{\left(u - 1 \right)}}{u} d u}}} = {\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)} - \sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)} - \sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}} = {\color{red}{\int{\left(\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} - \frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u}\right)d u}}}$$

Gliedweise integrieren:

$${\color{red}{\int{\left(\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} - \frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u}\right)d u}}} = {\color{red}{\left(- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} d u}\right)}}$$

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

$$- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + {\color{red}{\int{\frac{\sin{\left(u \right)} \cos{\left(1 \right)}}{u} d u}}} = - \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + {\color{red}{\cos{\left(1 \right)} \int{\frac{\sin{\left(u \right)}}{u} d u}}}$$

Dieses Integral (Sinusintegral) besitzt keine geschlossene Form:

$$- \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \cos{\left(1 \right)} {\color{red}{\int{\frac{\sin{\left(u \right)}}{u} d u}}} = - \int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u} + \cos{\left(1 \right)} {\color{red}{\operatorname{Si}{\left(u \right)}}}$$

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

$$\cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - {\color{red}{\int{\frac{\sin{\left(1 \right)} \cos{\left(u \right)}}{u} d u}}} = \cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - {\color{red}{\sin{\left(1 \right)} \int{\frac{\cos{\left(u \right)}}{u} d u}}}$$

Dieses Integral (Kosinusintegral) besitzt keine geschlossene Form:

$$\cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - \sin{\left(1 \right)} {\color{red}{\int{\frac{\cos{\left(u \right)}}{u} d u}}} = \cos{\left(1 \right)} \operatorname{Si}{\left(u \right)} - \sin{\left(1 \right)} {\color{red}{\operatorname{Ci}{\left(u \right)}}}$$

Zur Erinnerung: $$$u=x + 1$$$:

$$- \sin{\left(1 \right)} \operatorname{Ci}{\left({\color{red}{u}} \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left({\color{red}{u}} \right)} = - \sin{\left(1 \right)} \operatorname{Ci}{\left({\color{red}{\left(x + 1\right)}} \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left({\color{red}{\left(x + 1\right)}} \right)}$$

Daher,

$$\int{\frac{\sin{\left(x \right)}}{x + 1} d x} = - \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\sin{\left(x \right)}}{x + 1} d x} = - \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}+C$$

$$$\int \frac{\sin{\left(x \right)}}{x + 1}\, dx = \left(- \sin{\left(1 \right)} \operatorname{Ci}{\left(x + 1 \right)} + \cos{\left(1 \right)} \operatorname{Si}{\left(x + 1 \right)}\right) + C$$$A