参考にしたサイト➔ここだよ^^(No.7 ～ 12)

問題：次の式を計算せよ

• (1)の答え：\( \log|1 + \tan \dfrac{x}{2}| + C \)　(Cは積分定数とする)

  解説：

  \( t = \tan \dfrac{x}{2} \)とおくと、
  \( \sin x = \dfrac{2t}{1 + t^2}, \cos x = \dfrac{1 - t^2}{1 + t^2}, dx = \dfrac{2}{1 + t^2} dt \)

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


• (2)の答え：\( \dfrac{9}{128} \)

  解説：

  \( t = \sin x \)とおくと、
  \( dt = \cos xdx \)

  
  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{\frac{\pi}{3}} \sin^3 x\cos^3 x dx & = & \displaystyle \int_{0}^{\frac{\sqrt{3}}{2}} t^3\cos^2 x (\cos xdx) \\ & = &\displaystyle \int_{0}^{\frac{\sqrt{3}}{2}} t^3(1 - \sin^2 x) dt \\ & = & \displaystyle \int_{0}^{\frac{\sqrt{3}}{2}} t^3(1 - t^2) dt \\ & = & \displaystyle \int_{0}^{\frac{\sqrt{3}}{2}} t^3 - t^5 dt \\ & = & \left[ \dfrac{t^4}{4} - \dfrac{t^6}{6} \right]_{0}^{\frac{\sqrt{3}}{2}} \\ & = & \dfrac{9}{64} - \dfrac{9}{128} \\ & = & \dfrac{9}{128} \end{eqnarray} \)


• (3)の答え：\( \dfrac{2}{35} \)

  解説：

  \( t = \sqrt{1 - x^2} \)とおくと、 \( t^2 = 1 - x^2 \)
  \( tdt = -xdx \)

  
  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{1} (x\sqrt{1 - x^2})^3 dx & = & \displaystyle \int_{1}^{0} x^2t^3 x dx \\ & = & \displaystyle \int_{0}^{1} (1 - t^2)t^4 dt \\ & = & \displaystyle \int_{0}^{1} t^4 - t^6 dt \\ & = & \left[ \dfrac{t^5}{5} - \dfrac{t^7}{7} \right]_{0}^{1} \\ & = & \dfrac{1}{5} - \dfrac{1}{7} \\ & = & \dfrac{2}{35} \end{eqnarray} \)


• (4)の答え：\( \dfrac{1}{3} \)

  解説：

  \( I = \displaystyle \int_{-1}^{1} \dfrac{x^2}{1 + e^x} dx \) とおく。

  
  

  また、 \( \displaystyle \int_{-1}^{1} \dfrac{x^2}{1 + e^x} dx = \displaystyle \int_{-1}^{1} \dfrac{x^2}{1 + e^{-x}} dx \) より

  
  \( \hspace{35px} \begin{eqnarray} I & = & \dfrac{I + I}{2} \\ & = & \dfrac{1}{2}\displaystyle \int_{-1}^{1} \dfrac{x^2}{1 + e^x} + \dfrac{x^2}{1 + e^{-x}} dx \\ & = & \dfrac{1}{2}\displaystyle \int_{-1}^{1} \dfrac{x^2}{1 + e^x} + \dfrac{x^2e^x}{e^x + 1} dx \\ & = & \dfrac{1}{2}\displaystyle \int_{-1}^{1} \dfrac{x^2 + x^2e^x}{1 + e^x} dx \\ & = & \dfrac{1}{2}\displaystyle \int_{-1}^{1} \dfrac{x^2(1 + e^x)}{1 + e^x} dx \\ & = & \displaystyle \int_{0}^{1} x^2 dx \\ & = & \dfrac{1}{3} \\ \end{eqnarray} \)


• (5)の答え：\( \biggl( \dfrac{1}{2} - \dfrac{1}{\sqrt{3}} \biggr)\log 2 + \dfrac{\pi}{12} \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{1}^{\sqrt{3}} \dfrac{1}{x^2}\log\sqrt{1 + x^2} dx & = & \left[ -\dfrac{1}{x}\log\sqrt{1 + x^2} \right]_{1}^{\sqrt{3}} - \displaystyle \int_{1}^{\sqrt{3}} -\dfrac{1}{1 + x^2} dx \\ & = & -\dfrac{1}{\sqrt{3}}\log 2 + \dfrac{1}{2}\log 2 + \displaystyle \int_{1}^{\sqrt{3}} \dfrac{1}{1 + x^2} dx \\ & = & -\dfrac{1}{\sqrt{3}}\log 2 + \dfrac{1}{2}\log 2 + \left[ \theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\because x = \tan \thetaと置換した) \\ & = & \biggl( \dfrac{1}{2} - \dfrac{1}{\sqrt{3}} \biggr)\log 2 + (\dfrac{\pi}{4} - \dfrac{\pi}{6}) \\ & = & \biggl( \dfrac{1}{2} - \dfrac{1}{\sqrt{3}} \biggr)\log 2 + \dfrac{\pi}{12} \\ \end{eqnarray} \)