Integralen av $$$\sqrt{2} \cos{\left(2 x \right)}$$$

Bestäm $$$\int \sqrt{2} \cos{\left(2 x \right)}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\sqrt{2}$$$ och $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

$${\color{red}{\int{\sqrt{2} \cos{\left(2 x \right)} d x}}} = {\color{red}{\sqrt{2} \int{\cos{\left(2 x \right)} d x}}}$$

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,

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

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

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

Integralen av cosinus är $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

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

Kom ihåg att $$$u=2 x$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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