삼각치환,trig_substitution (rev. 1.6)
삼각 치환, trigonometric substitution
적분의 테크닉으로 쓰임. (삼각치환적분, integration by trigonometric substitution, ITS)
적분의 테크닉으로 쓰임. (삼각치환적분, integration by trigonometric substitution, ITS)
$\displaystyle \sqrt{a^2-x^2}$
$\displaystyle x=a\sin\theta$
$\displaystyle x=a\sin\theta$
$\displaystyle \sqrt{a^2+x^2}$
$\displaystyle x=a\tan\theta$
$\displaystyle x=a\tan\theta$
$\displaystyle \sqrt{x^2-a^2}$
$\displaystyle x=a\sec\theta$
$\displaystyle x=a\sec\theta$
i.e. (Stewart: Table of Trigonometric Substitutions)
| 식 | 치환 | θ의 범위 | 항등식 |
| $\displaystyle \sqrt{a^2-x^2}$ | $\displaystyle x=a\sin\theta$ | $\displaystyle -\frac{\pi}2\le\theta\le\frac{\pi}2$ | $\displaystyle 1-\sin^2\theta=\cos^2\theta$ |
| $\displaystyle \sqrt{a^2+x^2}$ | $\displaystyle x=a\tan\theta$ | $\displaystyle -\frac{\pi}2<\theta<\frac{\pi}2$ | $\displaystyle 1+\tan^2\theta=\sec^2\theta$ |
| $\displaystyle \sqrt{x^2-a^2}$ | $\displaystyle x=a\sec\theta$ | $\displaystyle 0\le\theta<\frac{\pi}2\textrm{ or }\pi\le\theta<\frac{3\pi}2$ | $\displaystyle \sec^2\theta-1=\tan^2\theta$ |
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
$\displaystyle x=\sec\theta(\pi<\theta<\frac32\pi)$
$\displaystyle dx=\sec\theta\tan\theta d\theta$
$\displaystyle x=-2\Rightarrow \cos\theta=-\frac12, \theta=\frac43\pi$
$\displaystyle x=-\sqrt{2}\Rightarrow \cos\theta=-\frac1{\sqrt{2}},\theta=\frac54\pi$
$\displaystyle \int_{\frac43\pi}^{\frac54\pi}\frac{\sec\theta\tan\theta}{\sqrt{\sec^2\theta-1}}d\theta$
$\displaystyle \int_{-2}^{-\sqrt{2}}\frac1{\sqrt{x^2-1}}dx$ 를 구하라.
sol.$\displaystyle x=\sec\theta(\pi<\theta<\frac32\pi)$
$\displaystyle dx=\sec\theta\tan\theta d\theta$
$\displaystyle x=-2\Rightarrow \cos\theta=-\frac12, \theta=\frac43\pi$
$\displaystyle x=-\sqrt{2}\Rightarrow \cos\theta=-\frac1{\sqrt{2}},\theta=\frac54\pi$
$\displaystyle \int_{\frac43\pi}^{\frac54\pi}\frac{\sec\theta\tan\theta}{\sqrt{\sec^2\theta-1}}d\theta$
