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

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

Skriv om integranden med hjälp av dubbelvinkelformeln $$$\sin\left(x \right)\cos\left(x \right)=\frac{1}{2}\sin\left( 2 x \right)$$$:

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

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

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

Använd potensreduceringsformeln $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ med $$$\alpha=2 x$$$:

$$\frac{{\color{red}{\int{\sin^{2}{\left(2 x \right)} d x}}}}{4} = \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{4}$$

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

$$\frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{4} = \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(4 x \right)}\right)d x}}{2}\right)}}}{4}$$

Integrera termvis:

$$\frac{{\color{red}{\int{\left(1 - \cos{\left(4 x \right)}\right)d x}}}}{8} = \frac{{\color{red}{\left(\int{1 d x} - \int{\cos{\left(4 x \right)} d x}\right)}}}{8}$$

Tillämpa konstantregeln $$$\int c\, dx = c x$$$ med $$$c=1$$$:

$$- \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{1 d x}}}}{8} = - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{x}}}{8}$$

Låt $$$u=4 x$$$ vara.

Då $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (stegen kan ses »), och vi har att $$$dx = \frac{du}{4}$$$.

Integralen blir

$$\frac{x}{8} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{8} = \frac{x}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{4}$$$ och $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{x}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{x}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$

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

$$\frac{x}{8} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{x}{8} - \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$

Kom ihåg att $$$u=4 x$$$:

$$\frac{x}{8} - \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{x}{8} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{32}$$

Alltså,

$$\int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}$$

Lägg till integrationskonstanten:

$$\int{\sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} d x} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}+C$$

$$$\int \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = \left(\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32}\right) + C$$$A