Integralen av $$$\cos{\left(\ln\left(11 x\right) \right)}$$$

Bestäm $$$\int \cos{\left(\ln\left(11 x\right) \right)}\, dx$$$.

Låt $$$u=11 x$$$ vara.

Då $$$du=\left(11 x\right)^{\prime }dx = 11 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{11}$$$.

Integralen kan omskrivas som

$${\color{red}{\int{\cos{\left(\ln{\left(11 x \right)} \right)} d x}}} = {\color{red}{\int{\frac{\cos{\left(\ln{\left(u \right)} \right)}}{11} d u}}}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{11}$$$ och $$$f{\left(u \right)} = \cos{\left(\ln{\left(u \right)} \right)}$$$:

$${\color{red}{\int{\frac{\cos{\left(\ln{\left(u \right)} \right)}}{11} d u}}} = {\color{red}{\left(\frac{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}{11}\right)}}$$

För integralen $$$\int{\cos{\left(\ln{\left(u \right)} \right)} d u}$$$, använd partiell integration $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Låt $$$\operatorname{g}=\cos{\left(\ln{\left(u \right)} \right)}$$$ och $$$\operatorname{dv}=du$$$.

Då gäller $$$\operatorname{dg}=\left(\cos{\left(\ln{\left(u \right)} \right)}\right)^{\prime }du=- \frac{\sin{\left(\ln{\left(u \right)} \right)}}{u} du$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{1 d u}=u$$$ (stegen kan ses »).

Integralen blir

$$\frac{{\color{red}{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}}}{11}=\frac{{\color{red}{\left(\cos{\left(\ln{\left(u \right)} \right)} \cdot u-\int{u \cdot \left(- \frac{\sin{\left(\ln{\left(u \right)} \right)}}{u}\right) d u}\right)}}}{11}=\frac{{\color{red}{\left(u \cos{\left(\ln{\left(u \right)} \right)} - \int{\left(- \sin{\left(\ln{\left(u \right)} \right)}\right)d u}\right)}}}{11}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=-1$$$ och $$$f{\left(u \right)} = \sin{\left(\ln{\left(u \right)} \right)}$$$:

$$\frac{u \cos{\left(\ln{\left(u \right)} \right)}}{11} - \frac{{\color{red}{\int{\left(- \sin{\left(\ln{\left(u \right)} \right)}\right)d u}}}}{11} = \frac{u \cos{\left(\ln{\left(u \right)} \right)}}{11} - \frac{{\color{red}{\left(- \int{\sin{\left(\ln{\left(u \right)} \right)} d u}\right)}}}{11}$$

För integralen $$$\int{\sin{\left(\ln{\left(u \right)} \right)} d u}$$$, använd partiell integration $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Låt $$$\operatorname{g}=\sin{\left(\ln{\left(u \right)} \right)}$$$ och $$$\operatorname{dv}=du$$$.

Då gäller $$$\operatorname{dg}=\left(\sin{\left(\ln{\left(u \right)} \right)}\right)^{\prime }du=\frac{\cos{\left(\ln{\left(u \right)} \right)}}{u} du$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{1 d u}=u$$$ (stegen kan ses »).

Integralen blir

$$\frac{u \cos{\left(\ln{\left(u \right)} \right)}}{11} + \frac{{\color{red}{\int{\sin{\left(\ln{\left(u \right)} \right)} d u}}}}{11}=\frac{u \cos{\left(\ln{\left(u \right)} \right)}}{11} + \frac{{\color{red}{\left(\sin{\left(\ln{\left(u \right)} \right)} \cdot u-\int{u \cdot \frac{\cos{\left(\ln{\left(u \right)} \right)}}{u} d u}\right)}}}{11}=\frac{u \cos{\left(\ln{\left(u \right)} \right)}}{11} + \frac{{\color{red}{\left(u \sin{\left(\ln{\left(u \right)} \right)} - \int{\cos{\left(\ln{\left(u \right)} \right)} d u}\right)}}}{11}$$

Vi har kommit till en integral som vi redan har sett.

Således har vi erhållit följande enkla ekvation med avseende på integralen:

$$\frac{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}{11} = \frac{u \sin{\left(\ln{\left(u \right)} \right)}}{11} + \frac{u \cos{\left(\ln{\left(u \right)} \right)}}{11} - \frac{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}{11}$$

Löser vi den får vi att

$$\int{\cos{\left(\ln{\left(u \right)} \right)} d u} = \frac{u \left(\sin{\left(\ln{\left(u \right)} \right)} + \cos{\left(\ln{\left(u \right)} \right)}\right)}{2}$$

Alltså,

$$\frac{{\color{red}{\int{\cos{\left(\ln{\left(u \right)} \right)} d u}}}}{11} = \frac{{\color{red}{\left(\frac{u \left(\sin{\left(\ln{\left(u \right)} \right)} + \cos{\left(\ln{\left(u \right)} \right)}\right)}{2}\right)}}}{11}$$

Kom ihåg att $$$u=11 x$$$:

$$\frac{{\color{red}{u}} \left(\sin{\left(\ln{\left({\color{red}{u}} \right)} \right)} + \cos{\left(\ln{\left({\color{red}{u}} \right)} \right)}\right)}{22} = \frac{{\color{red}{\left(11 x\right)}} \left(\sin{\left(\ln{\left({\color{red}{\left(11 x\right)}} \right)} \right)} + \cos{\left(\ln{\left({\color{red}{\left(11 x\right)}} \right)} \right)}\right)}{22}$$

Alltså,

$$\int{\cos{\left(\ln{\left(11 x \right)} \right)} d x} = \frac{x \left(\sin{\left(\ln{\left(11 x \right)} \right)} + \cos{\left(\ln{\left(11 x \right)} \right)}\right)}{2}$$

Förenkla:

$$\int{\cos{\left(\ln{\left(11 x \right)} \right)} d x} = \frac{\sqrt{2} x \sin{\left(\ln{\left(x \right)} + \frac{\pi}{4} + \ln{\left(11 \right)} \right)}}{2}$$

Lägg till integrationskonstanten:

$$\int{\cos{\left(\ln{\left(11 x \right)} \right)} d x} = \frac{\sqrt{2} x \sin{\left(\ln{\left(x \right)} + \frac{\pi}{4} + \ln{\left(11 \right)} \right)}}{2}+C$$

$$$\int \cos{\left(\ln\left(11 x\right) \right)}\, dx = \frac{\sqrt{2} x \sin{\left(\ln\left(x\right) + \frac{\pi}{4} + \ln\left(11\right) \right)}}{2} + C$$$A