자연계열 II 3번

【3-1】 삼각함수의 덧셈정리에 의해$\displaystyle \cos\left(x-\dfrac\pi4\right)=\cos x\cos\dfrac\pi4+\sin x\sin\dfrac\pi4=\dfrac1{\sqrt2}(\sin x+\cos x)$이다.

【3-2】 모든 음이 아닌 정수 $\displaystyle n$에 대해, $\displaystyle I_n=\int_0^{\pi/2}\dfrac1{(\sin x+\cos x)^n}\,dx$이라 하면, 3-1과 치환적분에 의해$\displaystyle \begin{aligned}I_n&=\int_0^{\pi/2}\dfrac1{(\sin x+\cos x)^n}\,dx\\&=\int_0^{\pi/2}\dfrac1{\left(\sqrt2\cos\left(x-\dfrac\pi4\right)\right)^n}\,dx\\&=2^{-\frac n2}\int_0^{\pi/2}\sec^n\left(x-\dfrac\pi4\right)\,dx\\&=2^{-\frac n2}\int_{-\pi/4}^{\pi/4}\sec^nx\,dx\end{aligned}$이다.$\displaystyle 2$보다 큰 정수 $\displaystyle n$에 대해, 닫힌구간 $\displaystyle \left[-\dfrac\pi4,\dfrac\pi4\right]$에서 두 함수 $\displaystyle f(x)=\sec^{n-2}x$와 $\displaystyle g(x)=\tan x$이 미분가능하고 $\displaystyle f^{\prime}(x)=(n-2)\sec^{n-2}x\tan x$와 $\displaystyle g^{\prime}(x)=\sec^2x$이 연속이므로, 부분적분법과 삼각함수의 성질에 의해$\displaystyle \begin{aligned}I_n&=2^{-\frac n2}\int_{-\pi/4}^{\pi/4}\sec^nx\,dx\\&=2^{-\frac n2}\Biggl[\sec^{n-2}x\tan x\Biggr]_{-\pi/4}^{\pi/4}-2^{-\frac n2}\int_{-\pi/4}^{\pi/4}(n-2)\sec^{n-2}x\tan^2x\,dx\\&=1-2^{-\frac n2}\int_{-\pi/4}^{\pi/4}(n-2)\sec^nx\,dx+2^{-\frac n2}\int_{-\pi/4}^{\pi/4}(n-2)\sec^{n-2}x\,dx\\&=1-(n-2)I_n+\dfrac{n-2}2I_{n-2}\end{aligned}$이다. 정리하면,$\displaystyle I_n=\dfrac1{n-1}+\dfrac{n-2}{2n-2}I_{n-2}$이고 $\displaystyle I_2=2^{-1}\int_{-\pi/4}^{\pi/4}\sec^2x\,dx=2^{-1}\Biggl[\tan x\Biggr]_{-\pi/4}^{\pi/4}=1$이다.

$\displaystyle \begin{aligned}I_4&=\dfrac13+\dfrac{I_2}3=\dfrac23& I_6&=\dfrac15+\dfrac{2I_4}5=\dfrac7{15}\\I_8&=\dfrac17+\dfrac{3I_6}7=\dfrac{12}{35}& I_{10}&=\dfrac19+\dfrac{4I_8}9=\dfrac{83}{315}\end{aligned}$따라서 $\displaystyle \int_0^{\pi/2}\dfrac1{(\sin x+\cos x)^{10}}\,dx=\dfrac{83}{315}$이다.

【3-3】 $\displaystyle x=\dfrac\pi2-t$에 대해, 치환적분법을 적용하면,$\displaystyle \begin{aligned}&\int_0^{\pi/2}\dfrac{x}{(\sin x+\cos x)^{10}}\,dx\\&=\int_{\pi/2}^0\dfrac{-\dfrac\pi2+t}{\left(\sin\left(\dfrac\pi2-t\right)+\cos\left(\dfrac\pi2-t\right)\right)^{10}}\,dt\\&=\int_{\pi/2}^0\dfrac{-\dfrac\pi2+t}{(\sin t+\cos t)^{10}}\,dt\\&=\int_0^{\pi/2}\dfrac{\dfrac\pi2-t}{(\sin t+\cos t)^{10}}\,dt\\&=\int_0^{\pi/2}\dfrac{\dfrac\pi2-x}{(\sin x+\cos x)^{10}}\,dx\\&=\dfrac\pi2\int_0^{\pi/2}\dfrac1{(\sin x+\cos x)^{10}}\,dx-\int_0^{\pi/2}\dfrac{x}{(\sin x+\cos x)^{10}}\,dx\end{aligned}$이다. 따라서 $\displaystyle \int_0^{\pi/2}\dfrac{x}{(\sin x+\cos x)^{10}}\,dx=\dfrac\pi4\int_0^{\pi/2}\dfrac1{(\sin x+\cos x)^{10}}\,dx=\dfrac{83\pi}{1260}$이다.