$$$x$$$에 대한 $$$\sqrt{- x^{2} y^{2} + 4}$$$의 적분

$$$\int \sqrt{- x^{2} y^{2} + 4}\, dx$$$을(를) 구하시오.

$$$x=\frac{2 \sin{\left(u \right)}}{\left|{y}\right|}$$$라 하자.

따라서 $$$dx=\left(\frac{2 \sin{\left(u \right)}}{\left|{y}\right|}\right)^{\prime }du = \frac{2 \cos{\left(u \right)}}{\left|{y}\right|} du$$$ (풀이 과정은 »에서 볼 수 있습니다).

또한 $$$u=\operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}$$$가 성립한다.

따라서,

$$$\sqrt{- x^{2} y^{2} + 4} = \sqrt{4 - 4 \sin^{2}{\left( u \right)}}$$$

$$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$ 항등식을 사용하시오:

$$$\sqrt{4 - 4 \sin^{2}{\left( u \right)}}=2 \sqrt{1 - \sin^{2}{\left( u \right)}}=2 \sqrt{\cos^{2}{\left( u \right)}}$$$

$$$\cos{\left( u \right)} \ge 0$$$라고 가정하면, 다음을 얻습니다:

$$$2 \sqrt{\cos^{2}{\left( u \right)}} = 2 \cos{\left( u \right)}$$$

적분은 다음과 같이 됩니다

$${\color{red}{\int{\sqrt{- x^{2} y^{2} + 4} d x}}} = {\color{red}{\int{\frac{4 \cos^{2}{\left(u \right)}}{\left|{y}\right|} d u}}}$$

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=\frac{4}{\left|{y}\right|}$$$와 $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$에 적용하세요:

$${\color{red}{\int{\frac{4 \cos^{2}{\left(u \right)}}{\left|{y}\right|} d u}}} = {\color{red}{\left(\frac{4 \int{\cos^{2}{\left(u \right)} d u}}{\left|{y}\right|}\right)}}$$

멱 감소 공식 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$를 $$$\alpha= u $$$에 적용하세요:

$$\frac{4 {\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}}{\left|{y}\right|} = \frac{4 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{\left|{y}\right|}$$

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=\frac{1}{2}$$$와 $$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$에 적용하세요:

$$\frac{4 {\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{\left|{y}\right|} = \frac{4 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}}{\left|{y}\right|}$$

각 항별로 적분하십시오:

$$\frac{2 {\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{\left|{y}\right|} = \frac{2 {\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{\left|{y}\right|}$$

상수 법칙 $$$\int c\, du = c u$$$을 $$$c=1$$$에 적용하십시오:

$$\frac{2 \left(\int{\cos{\left(2 u \right)} d u} + {\color{red}{\int{1 d u}}}\right)}{\left|{y}\right|} = \frac{2 \left(\int{\cos{\left(2 u \right)} d u} + {\color{red}{u}}\right)}{\left|{y}\right|}$$

$$$v=2 u$$$라 하자.

그러면 $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$du = \frac{dv}{2}$$$임을 얻습니다.

적분은 다음과 같이 다시 쓸 수 있습니다.

$$\frac{2 \left(u + {\color{red}{\int{\cos{\left(2 u \right)} d u}}}\right)}{\left|{y}\right|} = \frac{2 \left(u + {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}\right)}{\left|{y}\right|}$$

상수배 법칙 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$을 $$$c=\frac{1}{2}$$$와 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$에 적용하세요:

$$\frac{2 \left(u + {\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}\right)}{\left|{y}\right|} = \frac{2 \left(u + {\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}\right)}{\left|{y}\right|}$$

코사인의 적분은 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{2 \left(u + \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{2}\right)}{\left|{y}\right|} = \frac{2 \left(u + \frac{{\color{red}{\sin{\left(v \right)}}}}{2}\right)}{\left|{y}\right|}$$

다음 $$$v=2 u$$$을 기억하라:

$$\frac{2 \left(u + \frac{\sin{\left({\color{red}{v}} \right)}}{2}\right)}{\left|{y}\right|} = \frac{2 \left(u + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{2}\right)}{\left|{y}\right|}$$

다음 $$$u=\operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}$$$을 기억하라:

$$\frac{2 \left(\frac{\sin{\left(2 {\color{red}{u}} \right)}}{2} + {\color{red}{u}}\right)}{\left|{y}\right|} = \frac{2 \left(\frac{\sin{\left(2 {\color{red}{\operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}}} \right)}}{2} + {\color{red}{\operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}}}\right)}{\left|{y}\right|}$$

따라서,

$$\int{\sqrt{- x^{2} y^{2} + 4} d x} = \frac{2 \left(\frac{\sin{\left(2 \operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)} \right)}}{2} + \operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}\right)}{\left|{y}\right|}$$

공식 $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$을 사용하여 식을 간단히 하십시오:

$$\int{\sqrt{- x^{2} y^{2} + 4} d x} = \frac{2 \left(\frac{x \sqrt{- \frac{x^{2} \left|{y}\right|^{2}}{4} + 1} \left|{y}\right|}{2} + \operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}\right)}{\left|{y}\right|}$$

더 단순화하십시오:

$$\int{\sqrt{- x^{2} y^{2} + 4} d x} = \frac{x \sqrt{- x^{2} y^{2} + 4}}{2} + \frac{2 \operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}}{\left|{y}\right|}$$

적분 상수를 추가하세요:

$$\int{\sqrt{- x^{2} y^{2} + 4} d x} = \frac{x \sqrt{- x^{2} y^{2} + 4}}{2} + \frac{2 \operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}}{\left|{y}\right|}+C$$

$$$\int \sqrt{- x^{2} y^{2} + 4}\, dx = \left(\frac{x \sqrt{- x^{2} y^{2} + 4}}{2} + \frac{2 \operatorname{asin}{\left(\frac{x \left|{y}\right|}{2} \right)}}{\left|{y}\right|}\right) + C$$$A