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➔・高校生・受験生への挑戦状！積分最難問ランダム13題 | 受験の月,
　・ルートx^2+a^2の積分計算の２通りの方法 | 高校数学の美しい物語

問題：次の式を計算せよ

• (1)の答え：\( \tan x + \dfrac{1}{3}\tan^3 x + C \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int \dfrac{1}{\cos^4 x} dx & = & \displaystyle \int \dfrac{1}{\cos^2 x}\dfrac{1}{\cos^2 x} dx \\ & = & \displaystyle \int \dfrac{1}{\cos^2 x}(1 + \tan^2 x) dx \\ & = & \displaystyle \int \dfrac{1}{\cos^2 x} dx + \displaystyle \int \dfrac{\tan^2 x}{\cos^2 x} dx \\ & = & \tan x + C + \displaystyle \int t^2 dt　(t = \tan x) \\ & = & \tan x + \dfrac{1}{3}t^3 + C \\ & = & \tan x + \dfrac{1}{3}\tan^3 x + C \end{eqnarray} \)


• (2)の答え：\( -\dfrac{1}{4\tan x} - \dfrac{1}{12\tan^3 x} + C \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int \dfrac{1}{(1 - \cos 2x)^2} dx & = & \displaystyle \int \dfrac{1}{(2\sin^2 x)^2} dx \\ & = & \dfrac{1}{4}\displaystyle \int \dfrac{1}{\sin^4 x} dx \\ & = & \dfrac{1}{4}\displaystyle \int \dfrac{\sin^2 x + \cos^2 x}{\sin^4 x} dx \\ & = & \dfrac{1}{4}\biggl( \displaystyle \int \dfrac{1}{\sin^2 x} dx + \displaystyle \int \dfrac{1}{\sin^2 x\tan^2 x} dx \biggr) \\ & = & \dfrac{1}{4}\biggl( \displaystyle \int \dfrac{1}{\tan^2 x}\dfrac{1}{\cos^2 x} dx + \displaystyle \int \dfrac{1}{\tan^4 x}\dfrac{1}{\cos^2 x} dx \biggr)　(分数に \dfrac{\cos^2 x}{\cos^2 x} を掛けた) \\ & = & \dfrac{1}{4}\biggl( \displaystyle \int \dfrac{1}{t^2} dt + \displaystyle \int \dfrac{1}{t^4} dt \biggr)　(t = \tan x) \\ & = & \dfrac{1}{4}\biggl( -\dfrac{1}{t} - \dfrac{1}{3t^3} \biggr) + C \\ & = & -\dfrac{1}{4\tan x} - \dfrac{1}{12\tan^3 x} + C \end{eqnarray} \)
  
  

  ※\( t = \tan x \)置換でもできるよ(\( t = \tan \dfrac{x}{2} \)のノリで行けるよ)



• (3)の答え：\( \log(x + \sqrt{x^2 + 1}) + C \)


• (4)の答え：\( \dfrac{1}{2}\{ \log(x + \sqrt{x^2 + 1}) + x\sqrt{x^2 + 1} \} + C \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int \sqrt{x^2 + 1} dx & = & \displaystyle \int \dfrac{x^2 + 1}{\sqrt{x^2 + 1}} dx \\ & = & \displaystyle \int \dfrac{1}{\sqrt{x^2 + 1}} dx + \displaystyle \int \dfrac{x \cdot x}{\sqrt{x^2 + 1}} dx \\ \displaystyle \int \sqrt{x^2 + 1} dx & = & \log(x + \sqrt{x^2 + 1}) + C + x\sqrt{x^2 + 1} - \displaystyle \int \sqrt{x^2 + 1} dx　※部分積分 \\ 2\displaystyle \int \sqrt{x^2 + 1} dx & = & \log(x + \sqrt{x^2 + 1}) + x\sqrt{x^2 + 1} + C \\ & = & \dfrac{1}{2}\{ \log(x + \sqrt{x^2 + 1}) + x\sqrt{x^2 + 1} \} + C \end{eqnarray} \)


• (5)の答え：\( \dfrac{1}{4\sqrt{2}}\log\biggl| \dfrac{\sqrt{2 - x} - \sqrt{2(2 + x)}}{\sqrt{2 - x} + \sqrt{2(2 + x)}} \biggr| + C \)

  解説：

  \( \sqrt{4 - x^2} = \sqrt{(2 - x)(2 + x)} = (2 + x)\sqrt{\dfrac{2 - x}{2 + x}} \)

  ここで\( t = \sqrt{\dfrac{2 - x}{2 + x}} \)とおくと、\( (2 + x)t^2 = (2 - x) \Leftrightarrow (1 + t^2)x = 2 - 2t^2 \Leftrightarrow x = \dfrac{2(1 - t^2)}{1 + t^2} \)
  \( x + 2 = \dfrac{4}{1 + t^2} \) より \( dx = \dfrac{-8t}{(1 + t^2)^2} dt \)
  \( \therefore \sqrt{4 - x^2} = \dfrac{4}{1 + t^2} \cdot t = \dfrac{4t}{1 + t^2},　2 + 3x = \dfrac{4(2 - t^2)}{1 + t^2} \)

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int \dfrac{1}{(2 + 3x)\sqrt{4 - x^2}} dx & = & \displaystyle \int \dfrac{1}{\dfrac{4(2 - t^2)}{1 + t^2} \cdot \dfrac{4t}{1 + t^2}} \dfrac{-8t}{(1 + t^2)^2} dt \\ & = & \displaystyle \int \dfrac{1}{\dfrac{16t(2 - t^2)}{(1 + t^2)^2}} \dfrac{-8t}{(1 + t^2)^2} dt \\ & = & -\dfrac{1}{2}\displaystyle \int \dfrac{1}{2 - t^2} dt \\ & = & \dfrac{1}{2}\displaystyle \int \dfrac{1}{2\sqrt{2}}\biggl( \dfrac{1}{t - \sqrt{2}} - \dfrac{1}{t + \sqrt{2}} \biggr) dt \\ & = & \dfrac{1}{4\sqrt{2}}(\log|t - \sqrt{2}| - \log|t + \sqrt{2}|) + C \\ & = & \dfrac{1}{4\sqrt{2}}\log\biggl| \dfrac{t - \sqrt{2}}{t + \sqrt{2}} \biggr| + C \\ & = & \dfrac{1}{4\sqrt{2}}\log\biggl| \dfrac{\sqrt{\dfrac{2 - x}{2 + x}} - \sqrt{2}}{\sqrt{\dfrac{2 - x}{2 + x}} + \sqrt{2}} \biggr| + C \\ & = & \dfrac{1}{4\sqrt{2}}\log\biggl| \dfrac{\sqrt{2 - x} - \sqrt{2(2 + x)}}{\sqrt{2 - x} + \sqrt{2(2 + x)}} \biggr| + C \end{eqnarray} \)