logo

1/(1+x^2)의 적분 📂보조정리

1/(1+x^2)의 적분

공식

  • 정적분

11+x2dx=π \int_{-\infty}^{\infty} \dfrac{1}{1+x^{2}}dx = \pi

  • 부정적분

11+x2dx=tan1x+C \int \dfrac{1}{1+x^{2}}dx = \tan^{-1} x + C

CC는 적분상수이다.

증명

정적분

x=tanθx = \tan \theta로 치환하자. 그러면 적분 범위는 π2π2\displaystyle \int_{-\infty}^{\infty} \to \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}이고, tan=sec2\tan ^{\prime} = \sec^{2}이므로 dx=sec2dθdx = \sec^{2} d\theta이다.

11+x2dx=π2π211+tan2θsec2θdθ=π2π211+sin2θcos2θsec2θdθ=π2π21cos2θ+sin2θcos2θsec2θdθ=π2π211cos2θsec2θdθ=π2π2cos2θsec2θdθ=π2π2cos2θ1cos2θdθ=π2π2dθ=π2(π2)=π \begin{align*} \int_{-\infty}^{\infty} \dfrac{1}{1+x^{2}}dx &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{1 + \tan^{2}\theta} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{1 + \frac{\sin^{2} \theta}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{\frac{\cos^{2} \theta + \sin^{2} \theta}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{\frac{1}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2} \theta \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2} \theta \dfrac{1}{\cos^{2} \theta} d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \\ &= \frac{\pi}{2} - (-\frac{\pi}{2}) \\ &= \pi \end{align*}

부정적분

마찬가지로 x=tanθx = \tan \theta로 치환하면,

11+x2dx=11+tan2θsec2θdθ=cos2θ1cos2θdθ=dθ=θ+C=tan1x+C \begin{align*} \int \dfrac{1}{1+x^{2}}dx &= \int \dfrac{1}{1 + \tan^{2}\theta} \sec^{2} \theta d\theta \\ &= \int \cos^{2} \theta \dfrac{1}{\cos^{2} \theta} d\theta \\ &= \int d\theta \\ &= \theta + C \\ &= \tan^{-1} x + C \end{align*}