$$$\frac{1}{2 \sqrt{1 - x^{2}}}$$$의 적분

$$$\int \frac{1}{2 \sqrt{1 - x^{2}}}\, dx$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=\frac{1}{2}$$$와 $$$f{\left(x \right)} = \frac{1}{\sqrt{1 - x^{2}}}$$$에 적용하세요:

$${\color{red}{\int{\frac{1}{2 \sqrt{1 - x^{2}}} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{\sqrt{1 - x^{2}}} d x}}{2}\right)}}$$

$$$\frac{1}{\sqrt{1 - x^{2}}}$$$의 적분은 $$$\int{\frac{1}{\sqrt{1 - x^{2}}} d x} = \operatorname{asin}{\left(x \right)}$$$:

$$\frac{{\color{red}{\int{\frac{1}{\sqrt{1 - x^{2}}} d x}}}}{2} = \frac{{\color{red}{\operatorname{asin}{\left(x \right)}}}}{2}$$

따라서,

$$\int{\frac{1}{2 \sqrt{1 - x^{2}}} d x} = \frac{\operatorname{asin}{\left(x \right)}}{2}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{2 \sqrt{1 - x^{2}}} d x} = \frac{\operatorname{asin}{\left(x \right)}}{2}+C$$

$$$\int \frac{1}{2 \sqrt{1 - x^{2}}}\, dx = \frac{\operatorname{asin}{\left(x \right)}}{2} + C$$$A