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2011年 京大第一問(2)
小問集合の2番として定積分問題が出題されました。
入試の積分問題一覧はこちら
問題
\(\displaystyle\int_{0}^{\frac{1}{2}} (x+1)\sqrt{1-2x^2}dx\)
解答
根号を外すように置換を考えます。
その1
\(x=\displaystyle\frac{1}{\sqrt{2}}\sin\theta\)とおく。
※積分範囲は\(0\)～\(\displaystyle\frac{\pi}{4}\)になる。
\(\displaystyle\int_{0}^{\frac{1}{2}} (x+1)\sqrt{1-2x^2}dx\)
\(=\displaystyle\int_{0}^{\frac{\pi}{4}} \biggl(\displaystyle\frac{1}{\sqrt{2}}\sin\theta+1\biggr)\cos x \times \displaystyle\frac{1}{\sqrt{2}}\cos\theta d\theta\)
\(=\displaystyle\frac{1}{2}\displaystyle\int_{0}^{\frac{\pi}{4}}\sin\theta\cos^2\theta d\theta+\displaystyle\frac{1}{\sqrt{2}}\displaystyle\int_{0}^{\frac{\pi}{4}}\cos^2\theta d\theta\)
\(=\displaystyle\frac{1}{2}\biggl[-\displaystyle\frac{1}{3}\cos^3\theta\biggr]_{0}^{\frac{\pi}{4}}+\displaystyle\frac{1}{\sqrt{2}}\biggl[\displaystyle\frac{1}{2}\theta+\displaystyle\frac{1}{4}\sin 2\theta\biggr]_{0}^{\frac{\pi}{4}}\)
※後半は半角公式から積分
\(=-\displaystyle\frac{1}{6}\biggl(\displaystyle\frac{1}{2\sqrt{2}}-1\biggr)+\displaystyle\frac{1}{\sqrt{2}}\biggl(\displaystyle\frac{\pi}{8}+\displaystyle\frac{1}{4}\biggr)\)
\(=\displaystyle\frac{\sqrt{2}}{12}+\displaystyle\frac{1}{6}+\displaystyle\frac{\sqrt{2}}{16}\pi\)
その2
だいたい同じだが、最初から2つの積分に分離することで少し計算を楽にできる。
\(\displaystyle\int_{0}^{\frac{1}{2}} (x+1)\sqrt{1-2x^2}dx\)
\(=\displaystyle\int_{0}^{\frac{1}{2}} x\sqrt{1-2x^2}dx+\displaystyle\int_{0}^{\frac{1}{2}} \sqrt{1-2x^2}dx\)
第一項
\(\displaystyle\int_{0}^{\frac{1}{2}} x\sqrt{1-2x^2}dx\)
\(=\biggl[-\displaystyle\frac{1}{6}(1-2x^2)^{\frac{3}{2}}\biggr]_{0}^{\frac{1}{2}}\)
\(=-\displaystyle\frac{\sqrt{2}}{24}+\displaystyle\frac{1}{6}\)
第二項
\(\displaystyle\int_{0}^{\frac{1}{2}} \sqrt{1-2x^2}dx\)
\(=\displaystyle\frac{1}{\sqrt{2}}\displaystyle\int_{0}^{\frac{\pi}{4}}\cos^2\theta d\theta\)
※\(x=\displaystyle\frac{1}{\sqrt{2}}\sin\theta\)とおいた。
\(=\displaystyle\frac{1}{\sqrt{2}}\biggl[\displaystyle\frac{1}{2}\theta+\displaystyle\frac{1}{4}\sin 2\theta\biggr]_{0}^{\frac{\pi}{4}}\)
\(=\displaystyle\frac{\sqrt{2}}{16}\pi+\displaystyle\frac{\sqrt{2}}{8}\)
まとめ
\(-\displaystyle\frac{\sqrt{2}}{24}+\displaystyle\frac{1}{6}+\displaystyle\frac{\sqrt{2}}{16}\pi+\displaystyle\frac{\sqrt{2}}{8}\)
\(=\displaystyle\frac{\sqrt{2}}{12}+\displaystyle\frac{1}{6}+\displaystyle\frac{\sqrt{2}}{16}\pi\)
