$$$\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)}$$$'nin integrali

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

$$$\sin\left(x \right)\sin\left(2 x \right)$$$ ifadesini $$$\alpha=x$$$ ve $$$\beta=2 x$$$ ile $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ formülünü kullanarak yeniden yazın:

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

İfadeyi genişletin:

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

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(x \right)} = \sin{\left(3 x \right)} \cos{\left(x \right)} - \sin{\left(3 x \right)} \cos{\left(3 x \right)}$$$ ile uygula:

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

Her terimin integralini alın:

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

İntegralin içindeki ifadeyi $$$\alpha=3 x$$$ ve $$$\beta=x$$$ ile birlikte $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ formülünü kullanarak yeniden yazın:

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

Sabit katsayı kuralı $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(x \right)} = \sin{\left(2 x \right)} + \sin{\left(4 x \right)}$$$ ile uygula:

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

Her terimin integralini alın:

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

$$$u=2 x$$$ olsun.

Böylece $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (adımlar » görülebilir) ve $$$dx = \frac{du}{2}$$$ elde ederiz.

İntegral şu hale gelir

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

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{2}$$$ ve $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ ile uygula:

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

Sinüsün integrali $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$- \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\sin{\left(4 x \right)} d x}}{4} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\sin{\left(4 x \right)} d x}}{4} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

Hatırlayın ki $$$u=2 x$$$:

$$- \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\sin{\left(4 x \right)} d x}}{4} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\sin{\left(4 x \right)} d x}}{4} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

$$$u=4 x$$$ olsun.

Böylece $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (adımlar » görülebilir) ve $$$dx = \frac{du}{4}$$$ elde ederiz.

O halde,

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(4 x \right)} d x}}}}{4} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{4}$$

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{4}$$$ ve $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ ile uygula:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{4} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{4}\right)}}}{4}$$

Sinüsün integrali $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{16} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{16}$$

Hatırlayın ki $$$u=4 x$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{16} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}{2} - \frac{\cos{\left({\color{red}{\left(4 x\right)}} \right)}}{16}$$

$$$u=\sin{\left(3 x \right)}$$$ olsun.

Böylece $$$du=\left(\sin{\left(3 x \right)}\right)^{\prime }dx = 3 \cos{\left(3 x \right)} dx$$$ (adımlar » görülebilir) ve $$$\cos{\left(3 x \right)} dx = \frac{du}{3}$$$ elde ederiz.

Dolayısıyla,

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(3 x \right)} d x}}}}{2} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{\frac{u}{3} d u}}}}{2}$$

Sabit katsayı kuralı $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$'i $$$c=\frac{1}{3}$$$ ve $$$f{\left(u \right)} = u$$$ ile uygula:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{\frac{u}{3} d u}}}}{2} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\left(\frac{\int{u d u}}{3}\right)}}}{2}$$

Kuvvet kuralını $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ $$$n=1$$$ ile uygulayın:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\int{u d u}}}}{6}=- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{6}=- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{6}$$

Hatırlayın ki $$$u=\sin{\left(3 x \right)}$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{u}}^{2}}{12} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16} - \frac{{\color{red}{\sin{\left(3 x \right)}}}^{2}}{12}$$

Dolayısıyla,

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x} = - \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}$$

İntegrasyon sabitini ekleyin:

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)} d x} = - \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \sin{\left(3 x \right)}\, dx = \left(- \frac{\sin^{2}{\left(3 x \right)}}{12} - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(4 x \right)}}{16}\right) + C$$$A