弘前大学(理系) 2023年　問題4-2


$(2)\ \ 次の定積分を求めよ。$
\[\hspace{3em} \int _{\scriptsize{-\cfrac{\pi}{4}}}^{\scriptsize{\cfrac{\pi}{3}}} \cfrac{x}{\cos ^2 x}dx\]


(2)


$(\tan x )'=\cfrac{1}{\cos ^2 x} \ \ だから部分積分法を用いて$

\begin{eqnarray*} & &\int _{\scriptsize{-\cfrac{\pi}{4}}}^{\scriptsize{\cfrac{\pi}{3}}} \cfrac{x}{\cos ^2 x}dx\\ \\ &=&\big [x \tan x \big] _{\scriptsize{-\cfrac{\pi}{4}}}^{\scriptsize{\cfrac{\pi}{3}}} - \int _{\scriptsize{-\cfrac{\pi}{4}}}^{\scriptsize{\cfrac{\pi}{3}}} \tan x dx\\ \\ &=&\cfrac{\pi}{3} \times \sqrt{3} - (- \cfrac{\pi}{4}) \times (-1) - \int _{\scriptsize{-\cfrac{\pi}{4}}}^{\scriptsize{\cfrac{\pi}{3}}} \cfrac{\sin x}{\cos x}dx\\ \\ &=&\cfrac{\sqrt{3}}{3}\pi - \cfrac{\pi}{4} - \int _{\scriptsize{-\cfrac{\pi}{4}}}^{\scriptsize{\cfrac{\pi}{3}}} \cfrac{(-\cos x)'}{\cos x}dx\\ \\ &=&\big(\cfrac{\sqrt{3}}{3} - \cfrac{1}{4}\big)\pi + \big[\log |\cos x |\big] _{\scriptsize{-\cfrac{\pi}{4}}}^{\scriptsize{\cfrac{\pi}{3}}} \\ \\ &=&\big(\cfrac{\sqrt{3}}{3} - \cfrac{1}{4}\big)\pi + \log \big(\cos \cfrac{\pi}{3}\big) - \log \big(\cos (-\cfrac{\pi}{4})\big) \\ \\ &=&\big(\cfrac{\sqrt{3}}{3} - \cfrac{1}{4}\big)\pi + \log \cfrac{1}{2}- \log \cfrac{1}{\sqrt{2}} \\ \\ &=&\big(\cfrac{\sqrt{3}}{3} - \cfrac{1}{4}\big)\pi - \log 2 + \log \sqrt{2} \\ \\ &=&\big(\cfrac{\sqrt{3}}{3} - \cfrac{1}{4}\big)\pi - \log 2 + \cfrac{1}{2}\log 2 \\ \\ &=&\big(\cfrac{\sqrt{3}}{3} - \cfrac{1}{4}\big)\pi - \cfrac{1}{2}\log 2 \\ \end{eqnarray*}

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