$$$\sin{\left(x \right)} + e$$$の積分

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

項別に積分せよ:

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

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

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

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

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

したがって、

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

積分定数を加える:

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

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