$$$\sin{\left(x \right)} - \pi$$$の積分
入力内容
$$$\int \left(\sin{\left(x \right)} - \pi\right)\, dx$$$ を求めよ。
解答
項別に積分せよ:
$${\color{red}{\int{\left(\sin{\left(x \right)} - \pi\right)d x}}} = {\color{red}{\left(- \int{\pi d x} + \int{\sin{\left(x \right)} d x}\right)}}$$
$$$c=\pi$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:
$$\int{\sin{\left(x \right)} d x} - {\color{red}{\int{\pi d x}}} = \int{\sin{\left(x \right)} d x} - {\color{red}{\pi x}}$$
正弦関数の不定積分は$$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$です:
$$- \pi x + {\color{red}{\int{\sin{\left(x \right)} d x}}} = - \pi x + {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$
したがって、
$$\int{\left(\sin{\left(x \right)} - \pi\right)d x} = - \pi x - \cos{\left(x \right)}$$
積分定数を加える:
$$\int{\left(\sin{\left(x \right)} - \pi\right)d x} = - \pi x - \cos{\left(x \right)}+C$$
解答
$$$\int \left(\sin{\left(x \right)} - \pi\right)\, dx = \left(- \pi x - \cos{\left(x \right)}\right) + C$$$A