Integralen av $$$\frac{\sin{\left(2 x \right)}}{3}$$$

Bestäm $$$\int \frac{\sin{\left(2 x \right)}}{3}\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=\frac{1}{3}$$$ och $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(2 x \right)}}{3} d x}}} = {\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{3}\right)}}$$

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,

$$\frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{3} = \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{3}$$

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

$$\frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{3} = \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{3}$$

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

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{6} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{6}$$

Kom ihåg att $$$u=2 x$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{6} = - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{6}$$

Alltså,

$$\int{\frac{\sin{\left(2 x \right)}}{3} d x} = - \frac{\cos{\left(2 x \right)}}{6}$$

Lägg till integrationskonstanten:

$$\int{\frac{\sin{\left(2 x \right)}}{3} d x} = - \frac{\cos{\left(2 x \right)}}{6}+C$$

$$$\int \frac{\sin{\left(2 x \right)}}{3}\, dx = - \frac{\cos{\left(2 x \right)}}{6} + C$$$A