자연계열 1번

1. $\displaystyle f'(x)=\frac{2x(x+\sqrt{a^2-x^2})-x^2\left(1-\frac{x}{\sqrt{a^2-x^2}}\right)}{(x+\sqrt{a^2-x^2})^2}=\frac{x(x\sqrt{a^2-x^2}+2a^2-x^2)}{(x+\sqrt{a^2-x^2})^2\sqrt{a^2-x^2}}$이고, 모든 $\displaystyle 0\le x\le a$에 대하여 $\displaystyle x\sqrt{a^2-x^2}+2a^2-x^2>0$이므로 $\displaystyle f'(0)=0$이다.따라서 임의의 $\displaystyle 0<x<a$에 대하여 $\displaystyle f'(x)>0$이고 $\displaystyle f(x)$는 $\displaystyle [0,a]$에서 연속이므로, 함수 $\displaystyle f(x)$는 단조증가함수이고 일대일 함수이다.

2. $\displaystyle I=\int_0^\pi xf(a\sin x)\,dx$라 하면 $\displaystyle I=\int_0^{\pi/2}xf(a\sin x)\,dx+\int_{\pi/2}^\pi xf(a\sin x)\,dx$이고$\displaystyle t=\pi-x$로 치환하여 적분하면$\displaystyle \int_{\pi/2}^\pi xf(a\sin x)\,dx=\int_{\pi/2}^0(\pi-t)f(a\sin(\pi-t))(-dt)=\int_0^{\pi/2}(\pi-t)f(a\sin t)\,dt$이므로 $\displaystyle I=\pi\int_0^{\pi/2}f(a\sin x)\,dx$이다.

3. 2번의 결과로부터$\displaystyle I=\pi\int_0^{\pi/2}f(a\sin x)\,dx=\pi\int_0^{\pi/2}\frac{a^2\sin^2t}{a\sin t+a\cos t}\,dt=a\pi\int_0^{\pi/2}\frac{\sin^2t}{\sin t+\cos t}\,dt$이고, 다시 $\displaystyle \theta=\frac\pi2-t$로 치환하여 적분하면$\displaystyle I=a\pi\int_0^{\pi/2}\frac{\sin^2t}{\sin t+\cos t}\,dt=a\pi\int_{\pi/2}^0\frac{\sin^2(\frac\pi2-\theta)}{\sin(\frac\pi2-\theta)+\cos(\frac\pi2-\theta)}(-d\theta)=a\pi\int_0^{\pi/2}\frac{\cos^2\theta}{\cos\theta+\sin\theta}\,d\theta$따라서 $\displaystyle 2I=a\pi\int_0^{\pi/2}\frac{\sin^2\theta+\cos^2\theta}{\sin\theta+\cos\theta}\,d\theta=a\pi\int_0^{\pi/2}\frac1{\sin\theta+\cos\theta}\,d\theta$이고,$\displaystyle I=\frac{a\pi}2\int_0^{\pi/2}\frac{d\theta}{\sin\theta+\cos\theta}=\frac{a\pi}{2\sqrt2}\int_0^{\pi/2}\frac1{\sin(\theta+\frac\pi4)}\,d\theta=\frac{a\pi}{2\sqrt2}\int_{\pi/4}^{3\pi/4}\frac1{\sin\theta}\,d\theta=\frac{a\pi}{2\sqrt2}\int_{\pi/4}^{3\pi/4}\frac{\sin\theta}{1-\cos^2\theta}\,d\theta$이다. 다시 $\displaystyle \cos\theta=x$로 치환하여 적분하면,$\displaystyle I=\frac{a\pi}{2\sqrt2}\int_{\pi/4}^{3\pi/4}\frac{\sin\theta}{1-\cos^2\theta}\,d\theta=\frac{a\pi}{2\sqrt2}\int_{1/\sqrt2}^{-1/\sqrt2}\frac{(-dx)}{1-x^2}=\frac{a\pi}{2\sqrt2}\int_{-1/\sqrt2}^{1/\sqrt2}\frac{dx}{1-x^2}$이다.한편 $\displaystyle \frac1{1-x^2}=\frac12\left(\frac1{x+1}-\frac1{x-1}\right)$이므로,$\displaystyle I=\frac{a\pi}{4\sqrt2}\int_{-1/\sqrt2}^{1/\sqrt2}\frac1{x+1}-\frac1{x-1}\,dx=\Biggl[\frac{a\pi}{4\sqrt2}\ln\left|\frac{x+1}{x-1}\right|\Biggr]_{-1/\sqrt2}^{1/\sqrt2}=\frac{a\pi}{\sqrt2}\ln(1+\sqrt2)$이다.