2024 MIT Integration Bee Finals 해설 ②

Tiebreakers

Problem 1

${\displaystyle \int \frac{dx}{\sqrt[4]{x^4+1}}}$

적분을 변형하면

$$\int \frac{dx}{\sqrt[4]{x^4+1}}=\int \frac{dx}{x\sqrt[4]{1+{\displaystyle \frac{1}{x^4}}}}=\int \frac{{\displaystyle \frac{1}{x^5}}}{{\displaystyle \frac{1}{x^4}}\sqrt[4]{1+{\displaystyle \frac{1}{x^4}}}}dx$$

이다. $1+x^{-4}=u^4$으로 치환하자. $-x^{-5}dx=u^{3}du$이므로

$$\begin{array}{rl}\int \frac{dx}{\sqrt[4]{x^4+1}}& =-\int \frac{u^3}{(u^4-1)u}du\\ & =-\frac{1}{2}\int \frac{(u^2-1)+(u^2+1)}{(u^2+1)(u^2-1)}du\\ & =-\frac{1}{2}[\int \frac{1}{u^2+1}du+\int \frac{1}{u^2-1}du]\\ & =-\frac{1}{2}({\tan }^{-1}u+\frac{1}{2}\log \frac{u-1}{u+1})+C\\ & =-\frac{1}{2}{\tan }^{-1}\left(\frac{\sqrt[4]{x^4+1}}{x}\right)-\frac{1}{4}\log \left(\frac{\sqrt[4]{x^4+1}-x}{\sqrt[4]{x^4+1}+x}\right)+C\\ & =\boxed{{\displaystyle \frac{1}{2}{\tan }^{-1}\left(\frac{x}{\sqrt[4]{x^4+1}}\right)+\frac{1}{4}\log \left(\frac{\sqrt[4]{x^4+1}+x}{\sqrt[4]{x^4+1}-x}\right)+C}}\end{array}$$

이다.

Problem 2

${\displaystyle {\int }_0^{2\pi }\frac{(\sin 2x-5\sin x)\sin x}{\cos 2x-10\cos x+13}dx}$

피적분함수를 정리하면

$$\begin{array}{rl}\frac{(\sin 2x-5\sin x)\sin x}{\cos 2x-10\cos x+13}& =\frac{(2\cos x-5){\sin }^{2}x}{2(\cos x-2)(\cos x-3)}\\ & =\frac{1}{2}(\frac{1-{\cos }^{2}x}{\cos x-2}+\frac{1-{\cos }^{2}x}{\cos x-3})\\ & =\frac{1}{2}(-\cos x-2-\frac{3}{\cos x-2}-\cos x-3-\frac{8}{\cos x-3})\end{array}$$

이다. $a>1$에 대하여 ${\displaystyle {\int }_0^{2\pi }\frac{1}{\cos x-a}dx}$를 구하자. $t=\tan (x/2)$로 치환하면

$$\begin{array}{rl}{\int }_0^{2\pi }\frac{1}{\cos x-a}dx& ={\int }_0^{\pi }\frac{1}{\cos x-a}dx+{\int }_{\pi }^{2\pi }\frac{1}{\cos x-a}dx\\ & ={\int }_0^{\mathrm{\infty }}\frac{1}{{\displaystyle \frac{1-t^2}{1+t^2}}-a}\cdot \frac{2}{1+t^2}dt+{\int }_{-\mathrm{\infty }}^0\frac{1}{{\displaystyle \frac{1-t^2}{1+t^2}}-a}\cdot \frac{2}{1+t^2}dt\\ & =-\frac{2}{a+1}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{t^2+{\displaystyle \frac{a-1}{a+1}}}dt\\ & =-\frac{2}{a+1}\sqrt{\frac{a+1}{a-1}}\cdot \pi =-\frac{2\pi }{\sqrt{a^2-1}}\end{array}$$

이므로

$$\begin{array}{rl}{\int }_0^{2\pi }\frac{(\sin 2x-5\sin x)\sin x}{\cos 2x-10\cos x+13}dx& =\frac{1}{2}{\int }_0^{2\pi }(-5-2\cos x-\frac{3}{\cos x-2}-\frac{8}{\cos x-3})dx\\ & =\boxed{{\displaystyle (2\sqrt{2}+\sqrt{3}-5)\pi }}\end{array}$$

이다.

Problem 3

${\displaystyle \int \sqrt{x^4-4x+3}dx}$

$$\int \sqrt{x^4-4x+3}dx=\int |x-1|\sqrt{x^2+2x+3}dx$$

이다. $x+1=\sqrt{2}\sinh u$로 치환하자. ${\displaystyle {\sinh }^{-1}x=\log (x+\sqrt{x^2+1})}$이므로 $x\ge 1$에 대하여

$$\begin{array}{rl}& \int (\sqrt{2}\sinh u-2)\cdot 2{\cosh }^{2}udu\\ & =2\sqrt{2}\cdot \frac{{\cosh }^{3}u}{3}-2\int (\cosh (2u)+1)du\\ & =2\sqrt{2}\cdot \frac{{\cosh }^{3}u}{3}-\sinh (2u)-2u+C\\ & =\frac{2\sqrt{2}}{3}{(\frac{(x+1{)}^2}{2}+1)}^{3/2}-2\cdot \frac{x+1}{\sqrt{2}}\sqrt{\frac{(x+1{)}^2}{2}+1}-2{\sinh }^{-1}\frac{x+1}{\sqrt{2}}+C\\ & =\boxed{{\displaystyle \frac{1}{3}{(x^2+2x+3)}^{3/2}-(x+1)\sqrt{x^2+2x+3}-2\log (x+1+\sqrt{x^2+2x+3})+C}}\end{array}$$

이다. $x<1$이면

$$\begin{array}{c}\int \sqrt{x^4-4x+3}dx=-\int (x-1)\sqrt{x^2+2x+3}dx\hfill \\ \hfill =\boxed{{\displaystyle -\frac{1}{3}{(x^2+2x+3)}^{3/2}+(x+1)\sqrt{x^2+2x+3}+2\log (x+1+\sqrt{x^2+2x+3})+C}}\end{array}$$

이다.

Problem 4

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\sin }^2(2^x){\cos }^2(3^x)(4{\cos }^2(2^x){(4{\cos }^2(3^x)-3)}^2-1)dx}$

$\sin (2x)=2\sin (x)\cos (x)$이고, $\cos (3x)=4{\cos }^3(x)-3\cos (x)$이다. 따라서 피적분함수는

$$\begin{array}{rl}& {\sin }^2(2^x){\cos }^2(3^x)(4{\cos }^2(2^x){(4{\cos }^2(3^x)-3)}^2-1)\\ & ={(2{\sin }^2(2^x){\cos }^2(2^x))}^2{(4{\cos }^3(3^x)-3\cos (3^x))}^2-{\sin }^2(2^x){\cos }^2(3^x)\\ & ={\sin }^2\left(2^{x+1}\right){\cos }^2\left(3^{x+1}\right)-{\sin }^2(2^x){\cos }^2(3^x).\end{array}$$

이다. $f(x)={\sin }^2(2^x){\cos }^2(3^x)$로 두면 $f(x)\le max\{2^x,1\}$이다. $a\in (0,\mathrm{\infty })$에 대하여

$${\int }_{-\mathrm{\infty }}^{a}f(x)dx\le {\int }_{-\mathrm{\infty }}^0{2}^{x}dx+{\int }_0^{a}dx<\mathrm{\infty }$$

이므로

$${\int }_{-\mathrm{\infty }}^a(f(x+1)-f(x))dx={\int }_{-\mathrm{\infty }}^{a}f(x+1)dx-{\int }_{-\mathrm{\infty }}^{a}f(x)dx={\int }_a^{a+1}f(x)dx.$$

이다. 따라서

$$\begin{array}{rl}\underset{a\to \mathrm{\infty }}{lim}{\int }_{-\mathrm{\infty }}^a(f(x+1)-f(x))dx& =\underset{a\to \mathrm{\infty }}{lim}{\int }_a^{a+1}f(x)dx\\ & =\underset{a\to \mathrm{\infty }}{lim}{\int }_a^{a+1}\frac{1-\cos \left(2^{x+1}\right)}{2}\cdot \frac{1+\cos (2\cdot 3^x)}{2}dx\\ & =\frac{1}{4}\underset{a\to \mathrm{\infty }}{lim}{\int }_a^{a+1}(1-\cos \left(2^{x+1}\right)+\cos (2\cdot 3^x)-\cos \left(2^{x+1}\right)\cos (2\cdot 3^x))dx\\ & =\boxed{{\displaystyle \frac{1}{4}}}\end{array}$$

이다. 이때, 코사인 적분 함수 $\mathrm{Ci}(x)$에 대하여

$$\underset{a\to \mathrm{\infty }}{lim}{\int }_a^{a+1}\cos (2^x)dx=\underset{a\to \mathrm{\infty }}{lim}\frac{1}{\log 2}{\int }_{2^a}^{2^{a+1}}\frac{\cos (u)}{u}du=\underset{a\to \mathrm{\infty }}{lim}\frac{\mathrm{Ci}\left(2^{a+1}\right)-\mathrm{Ci}(2^a)}{\log 2}=0$$

이므로 나머지 항의 적분이 0이다.

Problem 5

${\displaystyle {\int }_2^{\mathrm{\infty }}\frac{\lfloor x\rfloor x^2}{x^6-1}dx}$

적분을 풀면

$$\begin{array}{rl}{\int }_2^{\mathrm{\infty }}\frac{\lfloor x\rfloor x^2}{x^6-1}dx& =\sum _{k=2}^{\mathrm{\infty }}\frac{k}{3}{\int }_k^{k+1}\frac{3x^2}{(x^3+1)(x^3-1)}dx\\ & =\sum _{k=2}^{\mathrm{\infty }}\frac{k}{3}{\int }_{k^3}^{(k+1{)}^3}\frac{1}{(u+1)(u-1)}du\\ & =\sum _{k=2}^{\mathrm{\infty }}\frac{k}{3}{[\frac{1}{2}\log \frac{u-1}{u+1}]}_{k^3}^{(k+1{)}^3}\\ & =\frac{1}{6}\log \left(\prod _{k=2}^{\mathrm{\infty }}{\left[\frac{(k+1{)}^3-1}{k^3-1}\right]}^k{\left[\frac{k^3+1}{(k+1{)}^3+1}\right]}^k\right)\end{array}$$

이므로

$$\begin{array}{rl}\prod _{k=2}^{\mathrm{\infty }}{\left[\frac{(k+1{)}^3-1}{k^3-1}\right]}^k{\left[\frac{k^3+1}{(k+1{)}^3+1}\right]}^k& =\underset{n\to \mathrm{\infty }}{lim}\prod _{k=2}^n{\left[\frac{(k+1{)}^3-1}{k^3-1}\right]}^k{\left[\frac{k^3+1}{(k+1{)}^3+1}\right]}^k\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{{[(n+1{)}^3-1]}^n}{7}\cdot \frac{9}{{[(n+1{)}^3+1]}^n}\prod _{k=2}^n\frac{k^3+1}{k^3-1}\\ & =\frac{9}{7}\underset{n\to \mathrm{\infty }}{lim}{\left[\frac{(n+1{)}^3-1}{(n+1{)}^3+1}\right]}^n\prod _{k=2}^n\frac{(k+1)(k^2-k+1)}{(k-1)(k^2+k+1)}\\ & =\frac{9}{7}\underset{n\to \mathrm{\infty }}{lim}\prod _{k=2}^n\frac{k+1}{k-1}\prod _{k=2}^n\frac{(k-1{)}^2+(k-1)+1}{k^2+k+1}\\ & =\frac{9}{7}\underset{n\to \mathrm{\infty }}{lim}\frac{n(n+1)}{1\cdot 2}\cdot \frac{3}{n^2+n+1}\\ & =\frac{27}{14}\end{array}$$

이다. 따라서

$${\int }_2^{\mathrm{\infty }}\frac{\lfloor x\rfloor x^2}{x^6-1}dx=\boxed{{\displaystyle \frac{1}{6}\log \frac{27}{14}}}$$

이다.

Lightning Round

Problem 1

${\displaystyle {\int }_0^1({(1-x^{\frac{3}{2}})}^{\frac{3}{2}}-{(1-x^{\frac{2}{3}})}^{\frac{2}{3}})dx}$

$y={(1-x^{\frac{3}{2}})}^{\frac{3}{2}}$로 두면 $x={(1-y^{\frac{2}{3}})}^{\frac{2}{3}}$이다. $y$가 $x$에 대한 순감소함수이므로

$${\int }_0^1{(1-x^{\frac{3}{2}})}^{\frac{3}{2}}dx={\int }_0^1{(1-x^{\frac{2}{3}})}^{\frac{2}{3}}dx$$

이다. 따라서 구하려는 적분은 $\boxed{{\displaystyle 0}}$이다.

Problem 2

${\displaystyle \int {\left(\frac{x}{x-1}\right)}^{4}dx}$

직접 계산하면

$$\begin{array}{rl}\int {\left(\frac{x}{x-1}\right)}^{4}dx& =\int {(1+\frac{1}{x-1})}^{4}dx\\ & =\int (1+\frac{4}{x-1}+\frac{6}{(x-1{)}^2}+\frac{4}{(x-1{)}^3}+\frac{1}{(x-1{)}^4})dx\\ & =\boxed{{\displaystyle x+4\log |x-1|-\frac{6}{x-1}-\frac{2}{(x-1{)}^2}-\frac{1}{3(x-1{)}^3}+C}}\end{array}$$

이다.

Problem 3

${\displaystyle \int \frac{(\tan (1012x)+\tan (1013x))\cos (1012x)\cos (1013x)}{\cos (2025x)}dx}$

피적분함수의 분자가

$$\sin (1012x)\cos (1013x)+\cos (1012x)\sin (1013x)=\sin (2025x)$$

이므로

$$\begin{array}{rl}\int \frac{(\tan (1012x)+\tan (1013x))\cos (1012x)\cos (1013x)}{\cos (2025x)}dx& =\int \tan (2025x)dx\\ & =\boxed{{\displaystyle -\frac{\log |\cos (2025x)|}{2025}+C}}\end{array}$$

이다.