Integralen av $$$\sin{\left(\ln\left(2 x\right) \right)}$$$

Bestäm $$$\int \sin{\left(\ln\left(2 x\right) \right)}\, dx$$$.

Låt $$$u=2 x$$$ vara.

Då $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{2}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = \sin{\left(\ln{\left(u \right)} \right)}$$$:

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

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

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

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

Alltså,

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

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

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

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

Alltså,

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

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

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

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

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

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

Löser vi den får vi att

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

Alltså,

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

Kom ihåg att $$$u=2 x$$$:

$$\frac{{\color{red}{u}} \left(\sin{\left(\ln{\left({\color{red}{u}} \right)} \right)} - \cos{\left(\ln{\left({\color{red}{u}} \right)} \right)}\right)}{4} = \frac{{\color{red}{\left(2 x\right)}} \left(\sin{\left(\ln{\left({\color{red}{\left(2 x\right)}} \right)} \right)} - \cos{\left(\ln{\left({\color{red}{\left(2 x\right)}} \right)} \right)}\right)}{4}$$

Alltså,

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

Förenkla:

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

Lägg till integrationskonstanten:

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

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