$$$\sin{\left(x \right)} \sin{\left(2 x \right)}$$$の積分

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)}\, dx$$$ を求めよ。

公式 $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ を用い、$$$\alpha=x$$$ および $$$\beta=2 x$$$ を用いて被積分関数を書き換えよ:

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

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{2}$$$ と $$$f{\left(x \right)} = \cos{\left(x \right)} - \cos{\left(3 x \right)}$$$ に対して適用する:

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

項別に積分せよ:

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

$$$u=3 x$$$ とする。

すると $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{3}$$$ となります。

積分は次のようになります

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

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{3}$$$ と $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ に対して適用する:

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

余弦の積分は$$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

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

次のことを思い出してください $$$u=3 x$$$:

$$\frac{\int{\cos{\left(x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{6} = \frac{\int{\cos{\left(x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{\left(3 x\right)}} \right)}}{6}$$

余弦の積分は$$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$:

$$- \frac{\sin{\left(3 x \right)}}{6} + \frac{{\color{red}{\int{\cos{\left(x \right)} d x}}}}{2} = - \frac{\sin{\left(3 x \right)}}{6} + \frac{{\color{red}{\sin{\left(x \right)}}}}{2}$$

したがって、

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

積分定数を加える:

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

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