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