$$$- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}$$$の積分

$$$\int \left(- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}\right)d x}}} = {\color{red}{\left(- \int{\frac{\sin{\left(x \right)}}{12} d x} + \int{2 \cos{\left(2 x \right)} d x}\right)}}$$

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

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + {\color{red}{\int{2 \cos{\left(2 x \right)} d x}}} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + {\color{red}{\left(2 \int{\cos{\left(2 x \right)} d x}\right)}}$$

$$$u=2 x$$$ とする。

すると $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$（手順は»で確認できます）、$$$dx = \frac{du}{2}$$$ となります。

この積分は次のように書き換えられる

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + 2 {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$

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

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + 2 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + 2 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$

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

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + {\color{red}{\int{\cos{\left(u \right)} d u}}} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + {\color{red}{\sin{\left(u \right)}}}$$

次のことを思い出してください $$$u=2 x$$$:

$$- \int{\frac{\sin{\left(x \right)}}{12} d x} + \sin{\left({\color{red}{u}} \right)} = - \int{\frac{\sin{\left(x \right)}}{12} d x} + \sin{\left({\color{red}{\left(2 x\right)}} \right)}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{12}$$$ と $$$f{\left(x \right)} = \sin{\left(x \right)}$$$ に対して適用する:

$$\sin{\left(2 x \right)} - {\color{red}{\int{\frac{\sin{\left(x \right)}}{12} d x}}} = \sin{\left(2 x \right)} - {\color{red}{\left(\frac{\int{\sin{\left(x \right)} d x}}{12}\right)}}$$

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

$$\sin{\left(2 x \right)} - \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{12} = \sin{\left(2 x \right)} - \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{12}$$

したがって、

$$\int{\left(- \frac{\sin{\left(x \right)}}{12} + 2 \cos{\left(2 x \right)}\right)d x} = \sin{\left(2 x \right)} + \frac{\cos{\left(x \right)}}{12}$$

積分定数を加える:

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

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