삼각치환을 적분

     역삼각함수를 이용한 삼각치환 적분

 

$$\int_{x_1}^{x_2}\frac{1}{a^2 +x^2}dx=\frac{1}{a}\left (     \tan ^{-1} \frac{x_2}{a}-\tan ^{-1}\frac{x_1}{a} \right )$$

[증명] $x=\tan \theta$ 로 치환하면 $dx=\sec ^{2} \theta d \theta$

$x_{1}=\tan\theta_{1}$,  $x_{2}=\tan\theta_{2}$ 에서

$\theta_{1}=\tan^{-1} x_{1}$,  $\theta_{2}=\tan^{-1} x_{2}$ 이므로

\begin{flalign}
& \displaystyle\int_{x_1}^{x_2}\frac{1}{a^2 +x^2}dx=\int_{\theta_1}^{\theta_2}\displaystyle\frac{1}{a^2 +a^2 \tan^{2} \theta} \sec ^{2} \theta d\theta \\

&=\int_{\theta_1}^{\theta_2}\displaystyle\frac{1}{a^2( 1+ \tan^{2}) \theta}\sec ^{2} \theta d\theta \\

&=\int_{\theta_1}^{\theta_2}\displaystyle\frac{1}{a^2 \sec ^{2} \theta }\sec ^{2} \theta d\theta \\

&=\int_{\theta_1}^{\theta_2}\frac{1}{a^2 } d\theta= \frac{1}{a^2 } \left( \theta_{1} - \theta_{2} \right ) \\

&=\frac{1}{a}\left (     \tan ^{-1} \frac{x_2}{a}-\tan ^{-1}\frac{x_1}{a} \right ) &&
 
\end{flalign}

여기서 $\tan^{-1} x$는  $\tan x$의 역함수이다.

(예) $\tan ^{-1} \displaystyle\frac{\sqrt{3}}{3}=\frac{\pi}{6}$,     $\tan ^{-1} 1=\displaystyle\frac{\pi}{4}$,     $\tan ^{-1} \sqrt{3}=\displaystyle\frac{\pi}{3}$  

 

Example  1

$\displaystyle\int_{0}^{2} \frac{1}{4+x^2}dx$의 값은?

[풀이] $\displaystyle\frac{1}{2}\left (  \tan ^{-1} \frac{2}{2}-\tan ^{-1}\frac{0}{2} \right )=\displaystyle\frac{1}{2} \left ( \frac{\pi}{4}-0\right)=\frac{\pi}{8}$ 

 

 

Example 2

$\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^2 x}dx$의 값은?

[풀이] $\cos x = t$로 치환하면 $-\sin x dx = dt$에서

\begin{align}

& \displaystyle \int_{1}^{0} \frac{-1}{1+t^2} dt \\

&=\displaystyle \int_{0}^{1} \frac{1}{1+t^2} dt \\

&=\displaystyle\frac{1}{1}\left (  \tan ^{-1} \frac{1}{1}-\tan ^{-1}\frac{0}{1} \right ) \\

&=\displaystyle\frac{1}{1} \left ( \frac{\pi}{4}-0\right)=\frac{\pi}{4}

\end{align}

 

Example 3

$ \displaystyle \lim_{n \to \infty}\sum_{k=1}^{n} \frac{4n}{n^2 + k^2} $의 값을 구하여라.

[풀이] $n \to 1$,   $k \to x$,  $\displaystyle\int_{0}^{1} (\;\;)dx $  (이전 글 참고)

\begin{align}

&\displaystyle\int_{0}^{1}\frac{4}{1+x^2}dx \\

&=\displaystyle\frac{4}{1}\left (  \tan ^{-1} \frac{1}{1}-\tan ^{-1}\frac{0}{1} \right ) \\

&=\displaystyle 4\left ( \frac{\pi}{4}-0\right)=\pi

\end{align}

 

 

$$\int_{x_1}^{x_2}\frac{1}{\sqrt{a^2 -x^2}}dx=     \sin ^{-1} \frac{x_2}{a}-\sin ^{-1}\frac{x_1}{a}$$

[증명] $x=a\sin \theta \; \left ( -\displaystyle\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2} \right )$  $(a>0)$

로 치환하면 $dx=a\cos  \theta d \theta$ 이다.

$x_{1}=\sin \theta_{1}$,  $x_{2}=\sin \theta_{2}$ 에서

$\theta_{1}=\sin^{-1} x_{1}$,  $\theta_{2}=\sin^{-1} x_{2}$ 이므로

\begin{flalign}
& \displaystyle\int_{x_1}^{x_2}\frac{1}{\sqrt{a^2 -x^2}}dx \\

&=\int_{\theta_1}^{\theta_2}\displaystyle\frac{1}{\sqrt{a^2 -a^2  \sin^{2} \theta}} a\cos \theta d\theta \\

&=\int_{\theta_1}^{\theta_2}\displaystyle\frac{1}{\sqrt{a^2( 1- \sin^{2} \theta )}} a\cos \theta d\theta \\

&=\int_{\theta_1}^{\theta_2}\displaystyle\frac{1}{a \sqrt{ \cos ^{2} \theta }}a\cos \theta d\theta \\

&=\int_{\theta_1}^{\theta_2}1\;d\theta=  \theta_{1} - \theta_{2} \\

&=\sin ^{-1} \frac{x_2}{a}-\sin ^{-1}\frac{x_1}{a} &&
 
\end{flalign}

마찬가지로  $\sin^{-1} x$는  $\sin x$의 역함수이다.

(예) $\sin ^{-1} \displaystyle\frac{1}{2}=\frac{\pi}{6}$,     $\sin ^{-1}  \displaystyle\frac{\sqrt{2}}{2}=\displaystyle\frac{\pi}{4}$,    $\sin ^{-1}  \displaystyle\frac{\sqrt{3}}{2}=\displaystyle\frac{\pi}{3}$ 

 

 

Example 4

$ \displaystyle\int_{0}^{\frac{3}{2}}\frac{1}{\displaystyle\sqrt{9-x^2}}dx $ 의 값은?

[해설] $ \sin ^{-1} \displaystyle\frac{\displaystyle\frac{3}{2}}{3}-\sin ^{-1}\frac{0}{3}=\frac{\pi}{6}-0=\frac{\pi}{6} $

 

 

Example 5

$ \displaystyle\int_{0}^{1}\frac{1}{\displaystyle\sqrt{2-x^2}}dx $ 의 값은?

[해설]  $ \sin ^{-1} \displaystyle\frac{1}{\sqrt{2}}-\sin ^{-1}\frac{0}{\sqrt{2}}=\frac{\pi}{4}-0=\frac{\pi}{4} $

 

 

 

참고로 $y=\sqrt{a^2 -x^2}$ 은 반원이므로 

$$ \int_{-a}^{a}\sqrt{a^2 -x^2} dx = \frac{\pi}{2}a^2 $$

가 성립한다.