[mathjax]
積分問題72番
計算
今回は置換してやっていきます。
\(x=\tan \theta\)とおく。（この置き換えは分母の形から考えてやってみてます）
\(\displaystyle\int_{0}^{\frac{\pi}{4}}\displaystyle\frac{\log(1+\tan \theta)}{1+\tan ^2\theta}\cdot \displaystyle\frac{d\theta}{\cos^2 \theta}=\displaystyle\int_{0}^{\frac{\pi}{4}} \log(1+\tan \theta)d\theta\)
ここからの変形が\(\tan \theta\)を戻すとループしてしまうので行き詰まりかと思うかもしれませんが
\(\tan \theta=\displaystyle\frac{\sin \theta}{\cos \theta}\) の変形をします。
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log(\displaystyle\frac{\sin \theta+\cos \theta}{\cos \theta})d\theta\)\(=\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\sin \theta+\cos \theta) d\theta\)\(-\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\cos \theta) d\theta\)
まずは第一項を計算していきます。
第一項
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\sin \theta+\cos \theta) d\theta\)\(=\displaystyle\int_{0}^{\frac{\pi}{4}}\log \biggl(\sqrt 2\cos (\theta-\displaystyle\frac{\pi}{4})\biggr) d\theta\)
\(\cos \theta\)にまとめた理由は後で相殺させるのを見据えて（実際解くときは手探り状態ながらに、うまくいくと信じてやっていく）
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log \biggl(\sqrt 2\cos (\theta-\displaystyle\frac{\pi}{4})\biggr) d\theta=\)\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log \sqrt 2 d\theta\)\(+\displaystyle\int_{0}^{\frac{\pi}{4}} \log \biggl(\cos (\theta-\displaystyle\frac{\pi}{4})\biggr) d\theta\)
前半
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log \sqrt 2 d\theta\)\(=\displaystyle\frac{\pi}{4}\log \sqrt 2=\displaystyle\frac{\pi}{8}\log 2\)
後半
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log \biggl(\cos (\theta-\displaystyle\frac{\pi}{4})\biggr) d\theta\) で、\(\theta-\displaystyle\frac{\pi}{4}=-t\)とおく。（負号は範囲を変えないため）
\(\displaystyle\int_{\frac{\pi}{4}}^{0} \log \biggl(\cos (-t)\biggr) (-dt)=\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\cos \theta) d\theta\)
第一項まとめ
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\sin \theta+\cos \theta) d\theta\)\(=\displaystyle\frac{\pi}{8}\log 2+\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\cos \theta) d\theta\)
計算続き
\(\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\sin \theta+\cos \theta) d\theta\)\(-\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\cos \theta) d\theta\)
\(=\displaystyle\frac{\pi}{8}\log 2+\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\cos \theta) d\theta\)\(-\displaystyle\int_{0}^{\frac{\pi}{4}} \log (\cos \theta) d\theta=\displaystyle\frac{\pi}{8}\log 2\)
が答えとなります。
答え
\(\displaystyle\frac{\pi}{8}\log 2\)
