Integralen av $$$\sin{\left(2 x \right)} - \cos{\left(2 x \right)}$$$
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Din inmatning
Bestäm $$$\int \left(\sin{\left(2 x \right)} - \cos{\left(2 x \right)}\right)\, dx$$$.
Lösning
Integrera termvis:
$${\color{red}{\int{\left(\sin{\left(2 x \right)} - \cos{\left(2 x \right)}\right)d x}}} = {\color{red}{\left(\int{\sin{\left(2 x \right)} d x} - \int{\cos{\left(2 x \right)} d x}\right)}}$$
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,
$$\int{\sin{\left(2 x \right)} d x} - {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = \int{\sin{\left(2 x \right)} d x} - {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$
Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$\int{\sin{\left(2 x \right)} d x} - {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = \int{\sin{\left(2 x \right)} d x} - {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$
Integralen av cosinus är $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$\int{\sin{\left(2 x \right)} d x} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{2} = \int{\sin{\left(2 x \right)} d x} - \frac{{\color{red}{\sin{\left(u \right)}}}}{2}$$
Kom ihåg att $$$u=2 x$$$:
$$\int{\sin{\left(2 x \right)} d x} - \frac{\sin{\left({\color{red}{u}} \right)}}{2} = \int{\sin{\left(2 x \right)} d x} - \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$
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
$$- \frac{\sin{\left(2 x \right)}}{2} + {\color{red}{\int{\sin{\left(2 x \right)} d x}}} = - \frac{\sin{\left(2 x \right)}}{2} + {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}$$
Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:
$$- \frac{\sin{\left(2 x \right)}}{2} + {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}} = - \frac{\sin{\left(2 x \right)}}{2} + {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}$$
Integralen av sinus är $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:
$$- \frac{\sin{\left(2 x \right)}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{2} = - \frac{\sin{\left(2 x \right)}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{2}$$
Kom ihåg att $$$u=2 x$$$:
$$- \frac{\sin{\left(2 x \right)}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{2} = - \frac{\sin{\left(2 x \right)}}{2} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$
Alltså,
$$\int{\left(\sin{\left(2 x \right)} - \cos{\left(2 x \right)}\right)d x} = - \frac{\sin{\left(2 x \right)}}{2} - \frac{\cos{\left(2 x \right)}}{2}$$
Förenkla:
$$\int{\left(\sin{\left(2 x \right)} - \cos{\left(2 x \right)}\right)d x} = - \frac{\sqrt{2} \sin{\left(2 x + \frac{\pi}{4} \right)}}{2}$$
Lägg till integrationskonstanten:
$$\int{\left(\sin{\left(2 x \right)} - \cos{\left(2 x \right)}\right)d x} = - \frac{\sqrt{2} \sin{\left(2 x + \frac{\pi}{4} \right)}}{2}+C$$
Svar
$$$\int \left(\sin{\left(2 x \right)} - \cos{\left(2 x \right)}\right)\, dx = - \frac{\sqrt{2} \sin{\left(2 x + \frac{\pi}{4} \right)}}{2} + C$$$A