$$ \hspace{10pt} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos x} $$
【解答】
\begin{eqnarray}
（与式）& = & \int_{0}^{\frac{\pi}{4}} \frac{\cos x}{\cos^2 x}dx \\
& = & \int_{0}^{\frac{\pi}{4}} \frac{\cos x}{1 – \sin^2 x}dx \\
& = & \int_{0}^{\frac{\pi}{4}} \frac{\cos x}{(1 – \sin x)(1 + \sin x)}dx \\
& = & \int_{0}^{\frac{\pi}{4}} \frac{\cos x}{(1 – \sin x)(1 + \sin x)}dx \\
& = & \int_{0}^{\frac{\pi}{4}} \frac{1}{2} \left( \frac{\cos x}{1 – \sin x} + \frac{\cos x}{1 + \sin x} \right)dx \\
& = & \frac{1}{2} \left[ – \log | 1 – \sin x | + \log | 1 + \sin x | \right]_{0}^{\frac{\pi}{4}} \\
& = & \frac{1}{2} \left( – \log \left| 1- \frac{1}{\sqrt{2}} \right| + \log \left| 1+ \frac{1}{\sqrt{2}} \right| \right) \\
& = & \frac{1}{2} \log \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right) \\
& = & \frac{1}{2} \log ( \sqrt{2} + 1 )^{2} \\
& = & \log ( \sqrt{2} + 1 ) \cdots\cdots（答）
\end{eqnarray}
