Funktion $$$2 \sin{\left(2 x \right)} \sin{\left(3 x \right)}$$$ integraali

Määritä $$$\int 2 \sin{\left(2 x \right)} \sin{\left(3 x \right)}\, dx$$$.

Kirjoita $$$\sin\left(2 x \right)\sin\left(3 x \right)$$$ uudelleen kaavaa $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ käyttäen, kun $$$\alpha=2 x$$$ ja $$$\beta=3 x$$$:

$${\color{red}{\int{2 \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x}}} = {\color{red}{\int{\left(\cos{\left(x \right)} - \cos{\left(5 x \right)}\right)d x}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=\frac{1}{2}$$$ ja $$$f{\left(x \right)} = 2 \cos{\left(x \right)} - 2 \cos{\left(5 x \right)}$$$:

$${\color{red}{\int{\left(\cos{\left(x \right)} - \cos{\left(5 x \right)}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(2 \cos{\left(x \right)} - 2 \cos{\left(5 x \right)}\right)d x}}{2}\right)}}$$

Integroi termi kerrallaan:

$$\frac{{\color{red}{\int{\left(2 \cos{\left(x \right)} - 2 \cos{\left(5 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{2 \cos{\left(x \right)} d x} - \int{2 \cos{\left(5 x \right)} d x}\right)}}}{2}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ käyttäen $$$c=2$$$ ja $$$f{\left(x \right)} = \cos{\left(5 x \right)}$$$:

$$\frac{\int{2 \cos{\left(x \right)} d x}}{2} - \frac{{\color{red}{\int{2 \cos{\left(5 x \right)} d x}}}}{2} = \frac{\int{2 \cos{\left(x \right)} d x}}{2} - \frac{{\color{red}{\left(2 \int{\cos{\left(5 x \right)} d x}\right)}}}{2}$$

Olkoon $$$u=5 x$$$.

Tällöin $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$dx = \frac{du}{5}$$$.

Näin ollen,

$$\frac{\int{2 \cos{\left(x \right)} d x}}{2} - {\color{red}{\int{\cos{\left(5 x \right)} d x}}} = \frac{\int{2 \cos{\left(x \right)} d x}}{2} - {\color{red}{\int{\frac{\cos{\left(u \right)}}{5} d u}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{1}{5}$$$ ja $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{\int{2 \cos{\left(x \right)} d x}}{2} - {\color{red}{\int{\frac{\cos{\left(u \right)}}{5} d u}}} = \frac{\int{2 \cos{\left(x \right)} d x}}{2} - {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{5}\right)}}$$

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

$$\frac{\int{2 \cos{\left(x \right)} d x}}{2} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{5} = \frac{\int{2 \cos{\left(x \right)} d x}}{2} - \frac{{\color{red}{\sin{\left(u \right)}}}}{5}$$

Muista, että $$$u=5 x$$$:

$$\frac{\int{2 \cos{\left(x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{5} = \frac{\int{2 \cos{\left(x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{\left(5 x\right)}} \right)}}{5}$$

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

$$- \frac{\sin{\left(5 x \right)}}{5} + \frac{{\color{red}{\int{2 \cos{\left(x \right)} d x}}}}{2} = - \frac{\sin{\left(5 x \right)}}{5} + \frac{{\color{red}{\left(2 \int{\cos{\left(x \right)} d x}\right)}}}{2}$$

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

$$- \frac{\sin{\left(5 x \right)}}{5} + {\color{red}{\int{\cos{\left(x \right)} d x}}} = - \frac{\sin{\left(5 x \right)}}{5} + {\color{red}{\sin{\left(x \right)}}}$$

Näin ollen,

$$\int{2 \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x} = \sin{\left(x \right)} - \frac{\sin{\left(5 x \right)}}{5}$$

Lisää integrointivakio:

$$\int{2 \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x} = \sin{\left(x \right)} - \frac{\sin{\left(5 x \right)}}{5}+C$$

$$$\int 2 \sin{\left(2 x \right)} \sin{\left(3 x \right)}\, dx = \left(\sin{\left(x \right)} - \frac{\sin{\left(5 x \right)}}{5}\right) + C$$$A