【問題】定積分（2022年東大）

【解説】

定積分の問題です。ひたすら計算です。計算ミスに注意しましょう。計算過程で \( \displaystyle \frac{1}{\cos x} \) の積分が出てきます。求め方をマスターしておきましょう。

【解答】

（１）

\begin{eqnarray}
f(x) &=& (\cos x) \log ( \cos x ) – \cos x + \int_{0}^{x}(\cos t)\log(\cos t) dt \\
f^{\prime}(x) &=& ( – \sin x ) \log ( \cos x ) + \cos x ( – \sin x ) \frac{1}{\cos x} – ( – \sin x ) + ( \cos x ) \log ( \cos x ) \\
&=& ( \cos x – \sin x ) \log ( \cos x )
\end{eqnarray}
\( f^{\prime}(x)=0 \) とおくと
\begin{eqnarray}
\cos x – \sin x &=& 0 \\
\cos x &=& \sin x \\
∴x &=& \frac{\pi}{4} \\
\log ( \cos x ) &=& 0 \\
\cos x &=& 1 \\
∴x &=& 0
\end{eqnarray}
\begin{array}{c|ccccc}
\hline
x & 0 & \cdots & \frac{\pi}{4} & \cdots & \frac{\pi}{2} \\
\hline
f^{\prime}(x) & 0 & – & 0 & + & \\
\hline
f(x) & & \searrow & 極小 & \nearrow & \\
\hline
\end{array}
増減表より \( x=\frac{\pi}{4} \) のとき \( f(x) \) は極小かつ最小となる。

（２）

\begin{eqnarray}
f \left( \frac{\pi}{4} \right) &=& \left( \cos \frac{\pi}{4} \right) \log \left(\cos \frac{\pi}{4} \right) – \cos \frac{\pi}{4} \\
&& \hspace{15pt} + \int_{0}^{\frac{\pi}{4}} ( \cos t ) \log ( \cos t ) dt \\
&=& \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) – \frac{1}{\sqrt{2}} \\
&& \hspace{15pt} + \int_{0}^{\frac{\pi}{4}} ( \cos t ) \log ( \cos t ) dt \cdots ① \\
\end{eqnarray}
ここで
\begin{eqnarray}
&& \int_{0}^{\frac{\pi}{4}} ( \cos t ) \log ( \cos t ) dt \\
&& \hspace{5pt} = \int_{0}^{\frac{\pi}{4}} ( \sin t )^{\prime} \log ( \cos t ) dt \\
&& \hspace{5pt} = \left[ ( \sin t ) \log ( \cos t ) \right]_{0}^{\frac{\pi}{4}} – \int_{0}^{\frac{\pi}{4}} \sin t \cdot (- \sin t ) \frac{1}{\cos t} dt \\
&& \hspace{5pt} = \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) + \int_{0}^{\frac{\pi}{4}} \frac{\sin^2 t}{\cos t} dt \\
&& \hspace{5pt} = \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) + \int_{0}^{\frac{\pi}{4}} \left( \frac{1 – \cos^2 t}{\cos t } \right) dt \\
&& \hspace{5pt} = \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) + \int_{0}^{\frac{\pi}{4}} \left( \frac{1}{\cos t } – \cos t \right) dt \cdots ②
\end{eqnarray}
ここで
\begin{eqnarray}
\frac{1}{\cos t} &=& \frac{\cos t}{\cos^2 t} \\
&=& \frac{\cos t}{1-\sin^2 t} \\
&=& \frac{\cos t}{(1-\sin t)(1+\sin t)} \\
&=& \frac{1}{2} \left( \frac{\cos t}{1-\sin t} + \frac{\cos t}{1+\sin t} \right) \\
&=& \frac{1}{2} \left[ – \{ \log (1-\sin t) \}^{\prime} + \{ \log (1+\sin t) \}^{\prime} \right] \cdots ③
\end{eqnarray}
③を②に代入すると
\begin{eqnarray}
&& \int_{0}^{\frac{\pi}{4}} ( \cos t ) \log ( \cos t ) dt \\
&& \hspace{5pt} = \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) + \left[ -\frac{1}{2} \log ( 1-\sin t) + \frac{1}{2} \log ( 1+\sin t ) -\sin t \right]_{0}^{\frac{\pi}{4}} \\
&& \hspace{5pt} = \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) – \frac{1}{2} \log \left( 1 – \frac{1}{\sqrt{2}} \right) + \frac{1}{2} \log \left( 1 + \frac{1}{\sqrt{2}} \right) – \frac{1}{\sqrt{2}} \\
&& \hspace{5pt} = \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) – \frac{1}{2} \log \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right) – \frac{1}{\sqrt{2}} \\
&& \hspace{5pt} = \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) – \frac{1}{2} \log \frac{1}{(\sqrt{2}-1)^2} – \frac{1}{\sqrt{2}} \\
&& \hspace{5pt} = \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) + \log \left( \sqrt{2}-1 \right) – \frac{1}{\sqrt{2}} \cdots ④
\end{eqnarray}
④を①に代入すると
\begin{eqnarray}
f \left( \frac{\pi}{4} \right) &=& \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) – \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) + \log ( \sqrt{2}-1 ) – \frac{1}{\sqrt{2}} \\
&=& \frac{2}{\sqrt{2}} \log \left( \frac{1}{\sqrt{2}} \right) – \frac{2}{\sqrt{2}} + \log ( \sqrt{2}-1 ) \\
&=& \sqrt{2} \log 2^{-\frac{1}{2}} – \sqrt{2} + \log ( \sqrt{2}-1 ) \\
&=& – \frac{1}{\sqrt{2}} \log 2 – \sqrt{2} + \log ( \sqrt{2}-1 ) \cdots （答）
\end{eqnarray}