Funktion $$$\sin{\left(2 x \right)} - \cos{\left(2 x \right)}$$$ integraali

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

Integroi termi kerrallaan:

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

Olkoon $$$u=2 x$$$.

Tällöin $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dx = \frac{du}{2}$$$.

Siis,

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

Kosinin integraali on $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

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

Muista, että $$$u=2 x$$$:

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

Olkoon $$$u=2 x$$$.

Tällöin $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dx = \frac{du}{2}$$$.

Integraali muuttuu muotoon

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

Sinifunktion integraali on $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

Muista, että $$$u=2 x$$$:

$$- \frac{\sin{\left(2 x \right)}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{2} = - \frac{\sin{\left(2 x \right)}}{2} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$

Näin ollen,

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

Sievennä:

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

Lisää integrointivakio:

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

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