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

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

Låt $$$u=\sin{\left(3 x \right)}$$$ vara.

Då $$$du=\left(\sin{\left(3 x \right)}\right)^{\prime }dx = 3 \cos{\left(3 x \right)} dx$$$ (stegen kan ses »), och vi har att $$$\cos{\left(3 x \right)} dx = \frac{du}{3}$$$.

Integralen blir

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{3}$$$ och $$$f{\left(u \right)} = u$$$:

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

Tillämpa potensregeln $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=1$$$:

$$\frac{{\color{red}{\int{u d u}}}}{3}=\frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{3}=\frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{3}$$

Kom ihåg att $$$u=\sin{\left(3 x \right)}$$$:

$$\frac{{\color{red}{u}}^{2}}{6} = \frac{{\color{red}{\sin{\left(3 x \right)}}}^{2}}{6}$$

Alltså,

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

Lägg till integrationskonstanten:

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

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