Integralen av $$$- \frac{\sin{\left(2 x \right)}}{4}$$$

Bestäm $$$\int \left(- \frac{\sin{\left(2 x \right)}}{4}\right)\, dx$$$.

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=- \frac{1}{4}$$$ och $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:

$${\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)}}{4}\right)d x}}} = {\color{red}{\left(- \frac{\int{\sin{\left(2 x \right)} d x}}{4}\right)}}$$

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,

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

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

$$- \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4} = - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{4}$$

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

$$- \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

Kom ihåg att $$$u=2 x$$$:

$$\frac{\cos{\left({\color{red}{u}} \right)}}{8} = \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

Alltså,

$$\int{\left(- \frac{\sin{\left(2 x \right)}}{4}\right)d x} = \frac{\cos{\left(2 x \right)}}{8}$$

Lägg till integrationskonstanten:

$$\int{\left(- \frac{\sin{\left(2 x \right)}}{4}\right)d x} = \frac{\cos{\left(2 x \right)}}{8}+C$$

$$$\int \left(- \frac{\sin{\left(2 x \right)}}{4}\right)\, dx = \frac{\cos{\left(2 x \right)}}{8} + C$$$A