先進工学部2024年第３問の問題文全文
(1) 定積分 \(\displaystyle \int_0^{\frac{\pi}{3}}\cos^2{t}dt\) を計算すると\(,\)
\begin{align}\int_0^{\frac{\pi}{3}}\cos^2{t}dt=\frac{~\fbox{$\hskip0.4emア\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}{~\fbox{$\hskip0.4emイ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}~\pi+\frac{~\fbox{$\hskip0.4emウ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}{~\fbox{$\hskip0.4emエ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}~\sqrt{~\fbox{$\hskip0.4emオ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}\end{align}
となる.
(2) 定積分 \(\displaystyle \int_0^{\frac{\pi}{3}}\sin^3{t}dt\) を計算すると\(,\)
\begin{align}\int_0^{\frac{\pi}{3}}\sin^3{t}dt=\frac{\fbox{$\hskip0.4emカ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}}{~\fbox{$\hskip0.4emキク\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}\end{align}
となる.
(3) 関数 \(f(x),~g(x)\) が
\begin{align}f(x)=\cos{x}+\int_0^{\frac{\pi}{3}}g(t)\sin{t}dt\end{align}
\begin{align}g(x)=\sin^2{x}+\int_0^{\frac{\pi}{3}}f(t)\cos{t}dt\end{align}
を満たすとする. このとき\(,\) 関数 \(f(x)\) は
\begin{align}f(x)=\cos{x}+\frac{\fbox{$\hskip0.4emケコ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}}{~\fbox{$\hskip0.4emサシス\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}+\frac{~\fbox{$\hskip0.4emセソ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}{~\fbox{$\hskip0.4emタチ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}~\sqrt{~\fbox{$\hskip0.4emツ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}+\left(\frac{\fbox{$\hskip0.4emテ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}}{~\fbox{$\hskip0.4emトナ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}+\frac{\fbox{$\hskip0.4em二\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}}{~\fbox{$\hskip0.4emヌネ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}~\sqrt{~\fbox{$\hskip0.4emノ\hskip0.4em\Rule{0pt}{0.4em}{0.4em}$}~}\right)\pi\end{align}
と書ける.
(1) の解答〜半角の公式で次数下げ〜
\begin{align}\int_0^{\frac{\pi}{3}}\cos^2{t}dt=\int_0^{\frac{\pi}{3}}\frac{1+\cos{2t}}{2}dt\end{align}
\begin{align}=\biggl[\frac{1}{2}t+\frac{1}{4}\sin{2t}\biggr]_0^{\frac{\pi}{3}}=\frac{1}{6}\pi +\frac{1}{8}\sqrt{3}~~~~\cdots \fbox{答}\end{align}
ア：１ イ：６ ウ：１ エ：８ オ：３
(2) の解答〜微分接触型は置換せずに処理〜
微分接触型の積分
\begin{align}\int f(g(x))g^{\prime}(x)dx=F(x)+C~(~C~は積分定数~)\end{align}
ただし\(,\) \(F(x)\) は \(f(x)\) の原始関数の \(1\) つ
\begin{align}\int_0^{\frac{\pi}{3}}\sin^3{t}dt=\int_0^{\frac{\pi}{3}}\sin{t}(1-\cos^2{t})dt\end{align}
quandle
\(f(x)=1-x^2,~g(x)=\cos{x}\) とすると\(,\)
\begin{align}f(g(x))g^{\prime}(x)=-\sin{x}(1-\cos^2{x})\end{align}
なので\(,\) 微分接触型になっていますね！
\begin{align}=\biggl[-\cos{t}+\frac{1}{3}\cos^3{t}\biggr]_0^{\frac{\pi}{3}}\end{align}
quandle
\(\displaystyle F(x)=x-\frac{1}{3}x^3\) に \(g(x)=\cos{x}\) を代入すればいいだけなので置換する必要なしです！
\begin{align}=\left(-\frac{1}{2}+1\right)+\left(\frac{1}{24}-\frac{1}{3}\right)=\frac{5}{24}~~~~\cdots \fbox{答}\end{align}
(3) の解答〜定数部分を置換して連立を解く〜
\begin{align}A=\int_0^{\frac{\pi}{3}}g(t)\sin{t}dt,~B=\int_0^{\frac{pi}{3}}f(t)\cos{t}dt\end{align}
とおく.
\begin{align}A=\int_0^{\frac{\pi}{3}}(\sin^2{t}+B)\sin{t}dt\end{align}
\begin{align}=\int_0^{\frac{\pi}{3}}\sin^3{t}dt+B\int_0^{\frac{\pi}{3}}\sin{t}dt\end{align}
\begin{align}=\frac{5}{24}+B\biggl[-\cos{t}\biggr]_0^{\frac{\pi}{3}}=\frac{5}{24}+B\left(-\frac{1}{2}+1\right)\end{align}
\begin{align}A=\frac{1}{2}B+\frac{5}{24}~~~~\cdots ①\end{align}
quandle
(2) の結果が使えるので少しだけ計算を省略できます！
\begin{align}B=\int_0^{\frac{\pi}{3}}(\cos{t}+A)\cos{t}dt\end{align}
\begin{align}=\int_0^{\frac{\pi}{3}}\cos^2{t}dt+A\int_0^{\frac{\pi}{3}}\cos{t}dt\end{align}
\begin{align}=\frac{\pi}{6}+\frac{\sqrt{3}}{8}+A\biggl[\sin{t}\biggr]_0^{\frac{\pi}{3}}\end{align}
\begin{align}B=\frac{\sqrt{3}}{2}A+\frac{\pi}{6}+\frac{\sqrt{3}}{8}~~~~\cdots ②\end{align}
quandle
(1) の結果が使えて少し楽になりましたね！さて\(,\) あとは①と②を連立すればいいのですが見るからに大変そうですね…式は煩雑でもやるべきことは一文字消去です. 丁寧にやりましょう！
① を ② に代入して\(,\)
\begin{align}B=\frac{\sqrt{3}}{2}\left(\frac{1}{2}B+\frac{5}{24}\right)+\frac{\pi}{6}+\frac{\sqrt{3}}{8}\end{align}
\begin{align}\left(1-\frac{\sqrt{3}}{4}\right)B=\frac{11\sqrt{3}}{48}+\frac{\pi}{6}\end{align}
\begin{align}B=\frac{11\sqrt{3}+8\pi}{48}\times \frac{4}{4-\sqrt{3}}=\frac{11\sqrt{3}+8\pi}{12(4-\sqrt{3})}\end{align}
① に代入して\(,\)
\begin{align}A=\frac{11\sqrt{3}+8\pi}{24(4-\sqrt{3})}+\frac{5}{24}\end{align}
\begin{align}=\frac{8\pi +20+6\sqrt{3}}{24(4-\sqrt{3})}=\frac{4\pi +10+3\sqrt{3}}{12(4-\sqrt{3})}\end{align}
\begin{align}=\frac{(4\pi +10+3\sqrt{3})(4+\sqrt{3})}{156}\end{align}
\begin{align}=\frac{(10+3\sqrt{3})(4+\sqrt{3})}{156}+\frac{4(4+\sqrt{3})}{156}\pi \end{align}
\begin{align}=\frac{49}{156}+\frac{11}{78}\sqrt{3}+\left(\frac{4}{39}+\frac{1}{39}\sqrt{3}\right)\pi ~~~~\cdots \fbox{答}\end{align}
ケ：４ コ：９ サ：１ シ：５ ス：６ セ：１
ソ：１ タ：７ チ：８ ツ：３ テ：４ ト：３
ナ：９ 二：１ ヌ：３ ネ：９ ノ：３
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