参考にしたサイト➔これだよ

問題：次の式を計算せよ（問題の積分をIとおく）

• (1)の答え：\( \dfrac{\pi}{4} \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\sin x}{\sin x + \cos x} dx & = & \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\cos x}{\sin x + \cos x} dx \\ & = & \dfrac{1}{2}\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\sin x + \cos x}{\sin x + \cos x} dx \\ & = & \dfrac{\pi}{4} \end{eqnarray} \)


• (2)の答え：\( \dfrac{\pi}{8}\log 2 \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) dx & = & \displaystyle \int_{0}^{\frac{\pi}{4}} \log\biggl( 1 + \dfrac{1 - \tan x}{1 + \tan x} \biggr) dx \\ & = & \displaystyle \int_{0}^{\frac{\pi}{4}} \log\biggl( \dfrac{2}{1 + \tan x} \biggr) dx \\ & = & \displaystyle \int_{0}^{\frac{\pi}{4}} \log2 - \log(1 + \tan x) dx \\ I & = & \displaystyle \int_{0}^{\frac{\pi}{4}} \log2 - I \\ 2I& = & \dfrac{\pi}{4}\log 2 \\ & = & \dfrac{\pi}{8}\log 2 \\ \end{eqnarray} \)


• (3)の答え：\( \dfrac{\pi}{4} \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{e^x + 1} dx & = & \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{e^{-x} + 1} dx \\ 2I & = & \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x\biggl( \dfrac{1}{e^x + 1} + \dfrac{1}{e^{-x} + 1} \biggr) dx \\ I & = & \dfrac{1}{2}\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \\ & = & \dfrac{1}{2}\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1 - \cos 2x}{2} \\ & = & \dfrac{1}{2}\displaystyle \int_{0}^{\frac{\pi}{2}} 1 - \cos 2x \\ & = & \dfrac{1}{2}\left[ x - \dfrac{1}{2}\sin 2x \right]_{0}^{\frac{\pi}{2}} \\ & = & \dfrac{\pi}{4} \end{eqnarray} \)


• (4)の答え：\( \dfrac{\pi}{3} \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{\pi} x\cos^2 x\sin x dx & = & \displaystyle \int_{0}^{\pi} (\pi - x)\cos^2 x\sin x dx \\ 2I & = & \pi\displaystyle \int_{0}^{\pi} \cos^2 x\sin x dx \\ I & = & \dfrac{\pi}{2}\left[ -\dfrac{1}{3}\cos^3 x \right]_{0}^{\pi} \\ & = & \dfrac{\pi}{6} - (-\dfrac{\pi}{6}) \\ & = & \dfrac{\pi}{3} \end{eqnarray} \)


• (5)の答え：\( \dfrac{\pi^2}{4} \)

  解説：

  \( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{\pi} \dfrac{x\sin x}{1 + \cos^2 x} dx & = & \displaystyle \int_{0}^{\pi} \dfrac{(\pi - x)\sin x}{1 + \cos^2 x} dx \\ I & = & \dfrac{\pi}{2}\displaystyle \int_{0}^{\pi} \dfrac{\sin x}{1 + \cos^2 x} dx \\ & = & \dfrac{\pi}{2}\displaystyle \int_{-1}^{1} \dfrac{1}{1 + t^2} dt \\ & = & \pi\displaystyle \int_{0}^{1} \dfrac{1}{1 + t^2} dt \\ & = & \pi\displaystyle \int_{0}^{\frac{\pi}{4}} du \\ & = & \dfrac{\pi^2}{4} \end{eqnarray} \)