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$$$- \frac{1}{\sqrt{1 - x^{2}}}$$$の積分
入力内容
$$$\int \left(- \frac{1}{\sqrt{1 - x^{2}}}\right)\, dx$$$ を求めよ。
解答
定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=-1$$$ と $$$f{\left(x \right)} = \frac{1}{\sqrt{1 - x^{2}}}$$$ に対して適用する:
$${\color{red}{\int{\left(- \frac{1}{\sqrt{1 - x^{2}}}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{\sqrt{1 - x^{2}}} d x}\right)}}$$
$$$\frac{1}{\sqrt{1 - x^{2}}}$$$ の不定積分は $$$\int{\frac{1}{\sqrt{1 - x^{2}}} d x} = \operatorname{asin}{\left(x \right)}$$$ です:
$$- {\color{red}{\int{\frac{1}{\sqrt{1 - x^{2}}} d x}}} = - {\color{red}{\operatorname{asin}{\left(x \right)}}}$$
したがって、
$$\int{\left(- \frac{1}{\sqrt{1 - x^{2}}}\right)d x} = - \operatorname{asin}{\left(x \right)}$$
積分定数を加える:
$$\int{\left(- \frac{1}{\sqrt{1 - x^{2}}}\right)d x} = - \operatorname{asin}{\left(x \right)}+C$$
解答
$$$\int \left(- \frac{1}{\sqrt{1 - x^{2}}}\right)\, dx = - \operatorname{asin}{\left(x \right)} + C$$$A