$$$\sin{\left(x \right)} \sin{\left(2 x \right)} \tan{\left(1 \right)}$$$'nin integrali

Bulun: $$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \tan{\left(1 \right)}\, dx$$$.

$$$\sin\left(x \right)\sin\left(2 x \right)$$$ 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:

$${\color{red}{\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \tan{\left(1 \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \tan{\left(1 \right)} d x}}}$$

İfadeyi genişletin:

$${\color{red}{\int{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \tan{\left(1 \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\cos{\left(3 x \right)} \tan{\left(1 \right)}}{2}\right)d x}}}$$

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

$${\color{red}{\int{\left(\frac{\cos{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\cos{\left(3 x \right)} \tan{\left(1 \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\cos{\left(x \right)} \tan{\left(1 \right)} - \cos{\left(3 x \right)} \tan{\left(1 \right)}\right)d x}}{2}\right)}}$$

Her terimin integralini alın:

$$\frac{{\color{red}{\int{\left(\cos{\left(x \right)} \tan{\left(1 \right)} - \cos{\left(3 x \right)} \tan{\left(1 \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\cos{\left(x \right)} \tan{\left(1 \right)} d x} - \int{\cos{\left(3 x \right)} \tan{\left(1 \right)} d x}\right)}}}{2}$$

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

$$- \frac{\int{\cos{\left(3 x \right)} \tan{\left(1 \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(x \right)} \tan{\left(1 \right)} d x}}}}{2} = - \frac{\int{\cos{\left(3 x \right)} \tan{\left(1 \right)} d x}}{2} + \frac{{\color{red}{\tan{\left(1 \right)} \int{\cos{\left(x \right)} d x}}}}{2}$$

Kosinüsün integrali $$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$:

$$- \frac{\int{\cos{\left(3 x \right)} \tan{\left(1 \right)} d x}}{2} + \frac{\tan{\left(1 \right)} {\color{red}{\int{\cos{\left(x \right)} d x}}}}{2} = - \frac{\int{\cos{\left(3 x \right)} \tan{\left(1 \right)} d x}}{2} + \frac{\tan{\left(1 \right)} {\color{red}{\sin{\left(x \right)}}}}{2}$$

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

$$\frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{{\color{red}{\int{\cos{\left(3 x \right)} \tan{\left(1 \right)} d x}}}}{2} = \frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{{\color{red}{\tan{\left(1 \right)} \int{\cos{\left(3 x \right)} d x}}}}{2}$$

$$$u=3 x$$$ olsun.

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

Dolayısıyla,

$$\frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\tan{\left(1 \right)} {\color{red}{\int{\cos{\left(3 x \right)} d x}}}}{2} = \frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\tan{\left(1 \right)} {\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{2}$$

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

$$\frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\tan{\left(1 \right)} {\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}}{2} = \frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\tan{\left(1 \right)} {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}}}{2}$$

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

$$\frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\tan{\left(1 \right)} {\color{red}{\int{\cos{\left(u \right)} d u}}}}{6} = \frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\tan{\left(1 \right)} {\color{red}{\sin{\left(u \right)}}}}{6}$$

Hatırlayın ki $$$u=3 x$$$:

$$\frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\tan{\left(1 \right)} \sin{\left({\color{red}{u}} \right)}}{6} = \frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\tan{\left(1 \right)} \sin{\left({\color{red}{\left(3 x\right)}} \right)}}{6}$$

Dolayısıyla,

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \tan{\left(1 \right)} d x} = \frac{\sin{\left(x \right)} \tan{\left(1 \right)}}{2} - \frac{\sin{\left(3 x \right)} \tan{\left(1 \right)}}{6}$$

Sadeleştirin:

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \tan{\left(1 \right)} d x} = \frac{\left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \tan{\left(1 \right)}}{6}$$

İntegrasyon sabitini ekleyin:

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \tan{\left(1 \right)} d x} = \frac{\left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \tan{\left(1 \right)}}{6}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \tan{\left(1 \right)}\, dx = \frac{\left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \tan{\left(1 \right)}}{6} + C$$$A