問題：次の式を計算せよ

※Cは積分定数である。

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

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

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

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

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

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