자연계열 1번

1-1

제시문 <나>에서 $\displaystyle \tan x=t$로 놓으면 $\displaystyle \frac{dt}{dx}=\sec^2x$이고, $\displaystyle x=0$일 때 $\displaystyle t=0$, $\displaystyle x=\frac\pi4$일 때 $\displaystyle t=1$이므로

$\displaystyle \int_0^{\frac\pi4}\tan x\sec^2x\,dx=\int_0^1t\,dt=\Biggl[\frac12t^2\Biggr]_0^1=\frac12$

1-2

$\displaystyle \begin{aligned}a_1&=\int_0^{\frac\pi4}\tan x\,dx-\int_0^{\frac\pi4}\tan^2x\,dx+\int_0^{\frac\pi4}\tan^3x\,dx-\int_0^{\frac\pi4}\tan^4x\,dx\\&=\int_0^{\frac\pi4}\tan x(1+\tan^2x)\,dx-\int_0^{\frac\pi4}\tan^2x(1+\tan^2x)\,dx\\&=\int_0^{\frac\pi4}\tan x\sec^2x\,dx-\int_0^{\frac\pi4}\tan^2x\sec^2x\,dx\end{aligned}$

문항 1-1과 같이 $\displaystyle \tan x=t$로 치환하여 계산하면

$\displaystyle a_1=\frac12-\frac13$

$\displaystyle \begin{aligned}a_2&=\int_0^{\frac\pi4}\tan^2x\,dx-\int_0^{\frac\pi4}\tan^3x\,dx+\int_0^{\frac\pi4}\tan^4x\,dx-\int_0^{\frac\pi4}\tan^5x\,dx\\&=\int_0^{\frac\pi4}\tan^2x(1+\tan^2x)\,dx-\int_0^{\frac\pi4}\tan^3x(1+\tan^2x)\,dx\\&=\int_0^{\frac\pi4}\tan^2x\sec^2x\,dx-\int_0^{\frac\pi4}\tan^3x\sec^2x\,dx\end{aligned}$

문항 1-1과 같이 $\displaystyle \tan x=t$로 치환하여 계산하면

$\displaystyle a_2=\frac13-\frac14$이므로 $\displaystyle a_1+a_2=\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)=\frac14$이다.

1-3

주어진 급수의 제$\displaystyle n$항까지의 부분합은 $\displaystyle \sum_{k=1}^na_k=I_1+I_3-I_{n+1}-I_{n+3}$이다. 따라서

$\displaystyle \begin{aligned}\sum_{n=1}^\infty a_n&=\lim_{n\to\infty}\left(\int_0^{\frac\pi4}\tan x\sec^2x\,dx-\int_0^{\frac\pi4}\tan^{n+1}x\sec^2x\,dx\right)\\&=\lim_{n\to\infty}\left(\frac12-\frac1{n+2}\right)\\&=\frac12\end{aligned}$