Integralen av $$$\cos^{2}{\left(8 x \right)}$$$

Bestäm $$$\int \cos^{2}{\left(8 x \right)}\, dx$$$.

Låt $$$u=8 x$$$ vara.

Då $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{8}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{8}$$$ och $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$:

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

Använd potensreduceringsformeln $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ med $$$\alpha= u $$$:

$$\frac{{\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}}{8} = \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{8}$$

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

$$\frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{8} = \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}}{8}$$

Integrera termvis:

$$\frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{16} = \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{16}$$

Tillämpa konstantregeln $$$\int c\, du = c u$$$ med $$$c=1$$$:

$$\frac{\int{\cos{\left(2 u \right)} d u}}{16} + \frac{{\color{red}{\int{1 d u}}}}{16} = \frac{\int{\cos{\left(2 u \right)} d u}}{16} + \frac{{\color{red}{u}}}{16}$$

Låt $$$v=2 u$$$ vara.

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

Integralen kan omskrivas som

$$\frac{u}{16} + \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{16} = \frac{u}{16} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{16}$$

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

$$\frac{u}{16} + \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{16} = \frac{u}{16} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{16}$$

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

$$\frac{u}{16} + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{32} = \frac{u}{16} + \frac{{\color{red}{\sin{\left(v \right)}}}}{32}$$

Kom ihåg att $$$v=2 u$$$:

$$\frac{u}{16} + \frac{\sin{\left({\color{red}{v}} \right)}}{32} = \frac{u}{16} + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{32}$$

Kom ihåg att $$$u=8 x$$$:

$$\frac{\sin{\left(2 {\color{red}{u}} \right)}}{32} + \frac{{\color{red}{u}}}{16} = \frac{\sin{\left(2 {\color{red}{\left(8 x\right)}} \right)}}{32} + \frac{{\color{red}{\left(8 x\right)}}}{16}$$

Alltså,

$$\int{\cos^{2}{\left(8 x \right)} d x} = \frac{x}{2} + \frac{\sin{\left(16 x \right)}}{32}$$

Lägg till integrationskonstanten:

$$\int{\cos^{2}{\left(8 x \right)} d x} = \frac{x}{2} + \frac{\sin{\left(16 x \right)}}{32}+C$$

$$$\int \cos^{2}{\left(8 x \right)}\, dx = \left(\frac{x}{2} + \frac{\sin{\left(16 x \right)}}{32}\right) + C$$$A