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

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

Skriv om integranden:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=16$$$ och $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$:

$${\color{red}{\int{16 \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\left(16 \int{\cos^{2}{\left(x \right)} d x}\right)}}$$

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

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

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

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

Integrera termvis:

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

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

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

Låt $$$u=2 x$$$ vara.

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

Integralen kan omskrivas som

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

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

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

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

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

Kom ihåg att $$$u=2 x$$$:

$$8 x + 4 \sin{\left({\color{red}{u}} \right)} = 8 x + 4 \sin{\left({\color{red}{\left(2 x\right)}} \right)}$$

Alltså,

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

Lägg till integrationskonstanten:

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

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