$$$\csc^{2}{\left(x \right)} + 1$$$の積分
入力内容
$$$\int \left(\csc^{2}{\left(x \right)} + 1\right)\, dx$$$ を求めよ。
解答
項別に積分せよ:
$${\color{red}{\int{\left(\csc^{2}{\left(x \right)} + 1\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\csc^{2}{\left(x \right)} d x}\right)}}$$
$$$c=1$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:
$$\int{\csc^{2}{\left(x \right)} d x} + {\color{red}{\int{1 d x}}} = \int{\csc^{2}{\left(x \right)} d x} + {\color{red}{x}}$$
$$$\csc^{2}{\left(x \right)}$$$ の不定積分は $$$\int{\csc^{2}{\left(x \right)} d x} = - \cot{\left(x \right)}$$$ です:
$$x + {\color{red}{\int{\csc^{2}{\left(x \right)} d x}}} = x + {\color{red}{\left(- \cot{\left(x \right)}\right)}}$$
したがって、
$$\int{\left(\csc^{2}{\left(x \right)} + 1\right)d x} = x - \cot{\left(x \right)}$$
積分定数を加える:
$$\int{\left(\csc^{2}{\left(x \right)} + 1\right)d x} = x - \cot{\left(x \right)}+C$$
解答
$$$\int \left(\csc^{2}{\left(x \right)} + 1\right)\, dx = \left(x - \cot{\left(x \right)}\right) + C$$$A