$$$1 - \sin{\left(x \right)}$$$の積分

$$$\int \left(1 - \sin{\left(x \right)}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(1 - \sin{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{1 d x} - \int{\sin{\left(x \right)} d x}\right)}}$$

$$$c=1$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$- \int{\sin{\left(x \right)} d x} + {\color{red}{\int{1 d x}}} = - \int{\sin{\left(x \right)} d x} + {\color{red}{x}}$$

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

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

したがって、

$$\int{\left(1 - \sin{\left(x \right)}\right)d x} = x + \cos{\left(x \right)}$$

積分定数を加える:

$$\int{\left(1 - \sin{\left(x \right)}\right)d x} = x + \cos{\left(x \right)}+C$$

$$$\int \left(1 - \sin{\left(x \right)}\right)\, dx = \left(x + \cos{\left(x \right)}\right) + C$$$A