次の定積分を求めよ （2019年京都大学 理系第１問2（１））
$$ \hspace{10pt} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{x}{\cos^2 x}dx $$
【解答】
\begin{eqnarray}
（与式）& = & \int_{0}^{\frac{\pi}{4}} x ( \tan x )^{\prime}dx \\
& = & \left[ x \tan x \right]_{0}^{\frac{\pi}{4}} – \int_{0}^{\frac{\pi}{4}} (x)^{\prime} \tan x dx \\
& = & \frac{\pi}{4} – \int_{0}^{\frac{\pi}{4}} \tan x dx
\end{eqnarray}
ここで
$$ \int \tan x dx = \int \frac{\sin x}{\cos x}dx = – \log | \cos x | + C $$
だから
\begin{eqnarray}
（与式）& = & \frac{\pi}{4} – \left[ – \log | \cos x | \right]_{0}^{\frac{\pi}{4}} \\
& = & \frac{\pi}{4} + \log \frac{1}{\sqrt{2}} \\
& = & \frac{\pi}{4} – \frac{1}{2} \log 2 \cdots\cdots （答）
\end{eqnarray}
