삼각 치환, trigonometric substitution
적분의 테크닉으로 쓰임. (삼각치환적분, integration by trigonometric substitution, ITS)
적분의 테크닉으로 쓰임. (삼각치환적분, integration by trigonometric substitution, ITS)
| 이런 꼴이면 | 이렇게 치환 |
| $\displaystyle \sqrt{a^2-x^2}$ | $\displaystyle x=a\sin\theta$ |
| $\displaystyle \sqrt{a^2+x^2}$ | $\displaystyle x=a\tan\theta$ |
| $\displaystyle \sqrt{x^2-a^2}$ | $\displaystyle x=a\sec\theta$ |
i.e. (Stewart: Table of Trigonometric Substitutions)
| 식 | 치환 | θ의 범위 | 항등식 |
| $\displaystyle \sqrt{a^2-x^2}$ | $\displaystyle x=a\sin\theta$ | $\displaystyle -\frac{\pi}2\le\theta\le\frac{\pi}2$ | $\displaystyle 1-\sin^2\theta=\cos^2\theta$ |
| $\displaystyle \sqrt{a^2+x^2}$ | $\displaystyle x=a\tan\theta$ | $\displaystyle -\frac{\pi}2<\theta<\frac{\pi}2$ | $\displaystyle 1+\tan^2\theta=\sec^2\theta$ |
| $\displaystyle \sqrt{x^2-a^2}$ | $\displaystyle x=a\sec\theta$ | $\displaystyle 0\le\theta<\frac{\pi}2\textrm{ or }\pi\le\theta<\frac{3\pi}2$ | $\displaystyle \sec^2\theta-1=\tan^2\theta$ |
셋째줄에서 θ의 범위를 저렇게 복잡하게 정한 이유는, 저 범위에서 $\displaystyle \tan\theta>0$ 이기 때문이다. 그래서 $\displaystyle a>0$ 일 때,
$\displaystyle a\sqrt{\sec^2\theta-1}$
$\displaystyle =a\sqrt{\tan^2\theta}$
이 단계에서 다음으로 넘어갈 수 있다.$\displaystyle =a\sqrt{\tan^2\theta}$
$\displaystyle =a\tan\theta$
(차영준, ![[http]](/wiki/imgs/http.png)
치환적분법 02 / 치환법칙 48분)예 ¶
Q:
$\displaystyle \int\frac1{1+x^2}dx=\int dt$
$\displaystyle \int\frac1{1+x^2}dx=\int dt$
Sol:
치환:
치환:
$\displaystyle t=\tan x$
$\displaystyle dt=\sec^2xdx$
식=$\displaystyle dt=\sec^2xdx$
$\displaystyle \int\frac1{1+\tan^2t}\sec^2$
TBW예
$\displaystyle x=\sec\theta(\pi<\theta<\frac32\pi)$
$\displaystyle dx=\sec\theta\tan\theta d\theta$
$\displaystyle x=-2\Rightarrow \cos\theta=-\frac12, \theta=\frac43\pi$
$\displaystyle x=-\sqrt{2}\Rightarrow \cos\theta=-\frac1{\sqrt{2}},\theta=\frac54\pi$
$\displaystyle \int_{-2}^{-\sqrt{2}}\frac1{\sqrt{x^2-1}}dx$ 를 구하라.
sol.$\displaystyle x=\sec\theta(\pi<\theta<\frac32\pi)$
$\displaystyle dx=\sec\theta\tan\theta d\theta$
$\displaystyle x=-2\Rightarrow \cos\theta=-\frac12, \theta=\frac43\pi$
$\displaystyle x=-\sqrt{2}\Rightarrow \cos\theta=-\frac1{\sqrt{2}},\theta=\frac54\pi$
$\displaystyle \int_{\frac43\pi}^{\frac54\pi}\frac{\sec\theta\tan\theta}{\sqrt{\sec^2\theta-1}}d\theta =\int_{\frac43\pi}^{\frac54\pi}\frac{\sec\theta\tan\theta}{\sqrt{\tan^2\theta}}d\theta =\int_{\frac43\pi}^{\frac54\pi}\frac{\sec\theta\tan\theta}{|\tan\theta|}d\theta$
이 영역 $\displaystyle (\pi<\theta<\frac32\pi)$ 에서는 $\displaystyle \tan\theta>0$ 이므로$\displaystyle =\int_{\frac43\pi}^{\frac54\pi}\frac{\sec\theta\tan\theta}{\tan\theta}d\theta =\int_{\frac43\pi}^{\frac54\pi}\sec\theta d\theta$
$\displaystyle =\left[\ln|\sec\theta+\tan\theta|\right]_{\frac43\pi}^{\frac54\pi}$
$\displaystyle =\ln\frac{\sqrt{2}-1}{2-\sqrt{3}}$
$\displaystyle =\left[\ln|\sec\theta+\tan\theta|\right]_{\frac43\pi}^{\frac54\pi}$
$\displaystyle =\ln\frac{\sqrt{2}-1}{2-\sqrt{3}}$
trigonometric_substitution
trigonometric_substitution ?
Trigonometric_substitution ?
Trigonometric_substitution ?
trigonometric substitution
trigonometric_substitution ?Trigonometric_substitution ?Trigonometric_substitution ?trigonometric substitution