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

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

$$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ の公式を用い、$$$\alpha=x$$$ および $$$\beta=2 x$$$ を用いて $$$\sin\left(x \right)\cos\left(2 x \right)$$$ を変形せよ:

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

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 $$$f{\left(x \right)} = 2 \sin{\left(x \right)} - 2 \sin{\left(3 x \right)}$$$ に対して適用する:

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

項別に積分せよ:

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

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

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

$$$u=3 x$$$ 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

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

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ に対して適用する:

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

正弦関数の不定積分は$$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$です:

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

次のことを思い出してください $$$u=3 x$$$:

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

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

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

正弦関数の不定積分は$$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$です:

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

したがって、

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

積分定数を加える:

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

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