Funktion $$$\sin{\left(x \right)} \sin{\left(3 x \right)}$$$ integraali

Määritä $$$\int \sin{\left(x \right)} \sin{\left(3 x \right)}\, dx$$$.

Kirjoita integraandi uudelleen käyttämällä kaavaa $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ yhdessä $$$\alpha=x$$$:n ja $$$\beta=3 x$$$:n kanssa:

$${\color{red}{\int{\sin{\left(x \right)} \sin{\left(3 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} - \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(x \right)} = \cos{\left(2 x \right)} - \cos{\left(4 x \right)}$$$:

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

Integroi termi kerrallaan:

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

Olkoon $$$u=4 x$$$.

Tällöin $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dx = \frac{du}{4}$$$.

Integraali muuttuu muotoon

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{1}{4}$$$ ja $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

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

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

Muista, että $$$u=4 x$$$:

$$\frac{\int{\cos{\left(2 x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{8} = \frac{\int{\cos{\left(2 x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{8}$$

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 voidaan kirjoittaa muotoon

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

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

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

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

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

Muista, että $$$u=2 x$$$:

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

Näin ollen,

$$\int{\sin{\left(x \right)} \sin{\left(3 x \right)} d x} = \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{8}$$

Sievennä:

$$\int{\sin{\left(x \right)} \sin{\left(3 x \right)} d x} = \sin^{3}{\left(x \right)} \cos{\left(x \right)}$$

Lisää integrointivakio:

$$\int{\sin{\left(x \right)} \sin{\left(3 x \right)} d x} = \sin^{3}{\left(x \right)} \cos{\left(x \right)}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(3 x \right)}\, dx = \sin^{3}{\left(x \right)} \cos{\left(x \right)} + C$$$A