Integral dari $$$\sin{\left(3 x \right)} \cos{\left(x \right)}$$$

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

Tulis ulang integran menggunakan rumus $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ dengan $$$\alpha=3 x$$$ dan $$$\beta=x$$$:

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

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(x \right)} = \sin{\left(2 x \right)} + \sin{\left(4 x \right)}$$$:

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

Integralkan suku demi suku:

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

Misalkan $$$u=2 x$$$.

Kemudian $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{2}$$$.

Integralnya menjadi

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

Integral dari sinus adalah $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

Ingat bahwa $$$u=2 x$$$:

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

Misalkan $$$u=4 x$$$.

Kemudian $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{4}$$$.

Oleh karena itu,

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

Integral dari sinus adalah $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

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

Ingat bahwa $$$u=4 x$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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