2021-M2-09

1. $\begin{array}{cl}
   & \dfrac{d}{d\theta} \ln (\sec\theta+\tan\theta) \\
   = & \dfrac{1}{\sec\theta+\tan\theta} \times (\sec\theta\tan\theta+\sec^2\theta) \\
   = & \dfrac{\sec\theta(\tan\theta+\sec\theta)}{\sec\theta+\tan\theta} \\\
   = & \sec\theta
   \end{array}$

2. 利用 (a)(i) 的結果，可得

   $\begin{array}{rcl}
   \sec\theta & = & \dfrac{d}{d\theta} \ln (\sec\theta+\tan\theta) \\
   \dint \sec\theta d\theta & = & \ln(\sec\theta+\tan\theta)+C
   \end{array}$

   由此，可得

   $\begin{array}{rcl}
   \dint \sec^3\theta d\theta & = & \dint \sec \theta \sec^2\theta d\theta \\
   \dint \sec^3\theta d\theta & = & \dint \sec \theta d\tan\theta \\
   \dint \sec^3\theta d\theta & = &\sec\theta\tan\theta-\dint\tan\theta d\sec\theta \text{ , 利用分部積分法}\\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta -\dint \tan\theta \times \sec\theta\tan\theta d \theta \\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta-\dint \tan^2\theta \sec\theta d\theta \\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta -\dint (\sec^2\theta -1)\sec\theta d\theta \\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta-\dint (\sec^3\theta -\sec\theta )d\theta \\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta-\dint \sec^3\theta d\theta +\dint\sec\theta d\theta \\
   2\dint \sec^3\theta d\theta & = & \sec\theta\tan\theta+\dint \sec\theta d\theta \\
   2\dint \sec^3\theta d\theta & = & \sec\theta\tan\theta+\ln(\sec\theta+\tan\theta)+C \\
   \dint \sec^3\theta d\theta & = & \dfrac{1}{2}\sec\theta\tan\theta+\dfrac{1}{2}\ln(\sec\theta+\tan\theta)+\dfrac{C}{2}
   \end{array}$

當 $x=-a$ 時，$u=a$。

當 $x=a$ 時，$u=-a$。

$\begin{array}{cl}
& \dint_{-a}^a g(x)h(x) dx \\
= & \dint_a^{-a}g(-u)h(-u) (-du) \\
= & \dint_{-a}^a g(-u)h(-u)du \\
= & \dint_{-a}^a g(-x)h(-x) dx \text{ , 其中 $u$ 為啞變數。}\\
= & \dint_{-a}^a g(-x)h(x) dx \text{ , 對於所有 $x\in\mathbb{R}$，$h(x)=h(-x)$。}
\end{array}$

由於對於所有 $x\in\mathbb{R}$，$h(x)=h(-x)$，則 $h(x)$ 為一偶函數。由此，可得

$\begin{array}{cl}
& \dint_0^a h(x) dx \\
= & \dfrac{1}{2} \dint_{-a}^a h(x) dx \\
= & \dfrac{1}{2} \dint_{-a}^a [g(x)+g(-x)] h(x) dx \text{ , 對於所有 $x\in\mathbb{R}$，$g(x)+g(-x)=1$。} \\
= & \dfrac{1}{2}\left(\dint_{-a}^ag(x)h(x) dx +\dint_{-a}^a g(-x)h(x)dx \right) \\
= & \dfrac{1}{2}\left(\dint_{-a}^a g(x)h(x)dx + \dint_{-a}^a g(x)h(x) dx\right) \text{ , 上面已證。} \\
= & \dfrac{1}{2} \times 2 \dint_{-a}^a g(x)h(x) dx \\
= & \dint_{-a}^a g(x)h(x) dx
\end{array}$

檢查：

$\begin{array}{cl}
& g(x) + g(-x) \\
= & \dfrac{3^x}{3^x+3^{-x}}+\dfrac{3^{-x}}{3^{-x}+3^{-(-x)}} \\
= & \dfrac{3^x}{3^x+3^{-x}}+\dfrac{3^{-x}}{3^{-x}+3^{x}} \\
= & \dfrac{3^x+3^{-x}}{3^x+3^{-x}} \\
= & 1
\end{array}$

另外，

$\begin{array}{cl}
& h(-x) \\
= & \dfrac{(-x)^2}{\sqrt{(-x)^2+1}} \\
= & \dfrac{x^2}{\sqrt{x^2+1}} \\
= & h(x)
\end{array}$

由此，我們可以在 (c) 部中利用 (b) 部的結果。對於 $a=1$，可得

$\begin{array}{cl}
& \dint_{-1}^1 \dfrac{3^x x^2}{(3^x+3^{-x})\sqrt{x^2+1}}dx \\
= & \dint_{-1}^1 g(x)h(x) dx \\
= & \dint_0^1 h(x)dx \text{ , 利用 (b) 的結果。} \\
= & \dint_0^1\dfrac{x^2}{\sqrt{x^2+1}}dx
\end{array}$

設 $x=\tan\theta$。則 $dx =\sec^2\theta d\theta$。

當 $x=0$ 時，$\theta=0$。

當 $x=1$ 時，$\theta=\dfrac{\pi}{4}$。

$\begin{array}{cl}
& \dint_0^1\dfrac{x^2}{\sqrt{x^2+1}}dx \\
= & \dint_0^\frac{\pi}{4} \dfrac{\tan^2\theta}{\sqrt{\tan^2\theta+1}} \times \sec^2\theta d\theta \\
= & \dint_0^\frac{\pi}{4} \dfrac{\tan^2\theta\sec^2\theta}{\sqrt{\sec^2\theta}} d\theta \\
= & \dint_0^\frac{\pi}{4}(\sec^2\theta-1)\sec\theta d\theta \\
= & \dint_0^\frac{\pi}{4} (\sec^3\theta-\sec\theta)d\theta \\
= & \left[\dfrac{1}{2}\sec\theta\tan\theta+\dfrac{1}{2}\ln(\sec\theta+\tan\theta)\right]_0^\frac{\pi}{4}-\left[\ln(\sec\theta+\tan\theta)\right]_0^\frac{\pi}{4} \text{ , 利用 (a)(ii) 的結果。} \\
= & \dfrac{1}{2}\sec \dfrac{\pi}{4}\tan\dfrac{\pi}{4}+\dfrac{1}{2}\ln\left(\sec\dfrac{\pi}{4}+\tan\dfrac{\pi}{4}\right) -\dfrac{1}{2}\sec 0\tan 0 +\dfrac{1}{2}\ln(\sec 0+ \tan 0) \\
& \ \ \ \ -\ln\left(\sec\dfrac{\pi}{4}+\tan\dfrac{\pi}{4}\right)+\ln(\sec 0+ \tan 0) \\
= & \dfrac{1}{2}\times \sqrt{2}\times 1+\dfrac{1}{2}\ln(\sqrt{2}+1)-\dfrac{1}{2}\times 1 \times 0+\dfrac{1}{2}\ln(1+0)-\ln(\sqrt{2}+1)+\ln(1+0) \\
= & \dfrac{\sqrt{2}}{2} -\dfrac{1}{2}\ln(\sqrt{2}+1)
\end{array}$