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. By the result of (a)(i), we have

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

   Hence, we have

   $\begin{array}{rcl}
   \dint \sec^3\theta d\theta & = & \dint \sec \theta \sec^2\theta d\theta \\
   \dint \sec^3\theta d\theta & = & \dint \sec \theta d\tan\theta \\
   \dint \sec^3\theta d\theta & = &\sec\theta\tan\theta-\dint\tan\theta d\sec\theta \text{ , by integration by parts.}\\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta -\dint \tan\theta \times \sec\theta\tan\theta d \theta \\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta-\dint \tan^2\theta \sec\theta d\theta \\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta -\dint (\sec^2\theta -1)\sec\theta d\theta \\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta-\dint (\sec^3\theta -\sec\theta )d\theta \\
   \dint \sec^3\theta d\theta & = & \sec\theta\tan\theta-\dint \sec^3\theta d\theta +\dint\sec\theta d\theta \\
   2\dint \sec^3\theta d\theta & = & \sec\theta\tan\theta+\dint \sec\theta d\theta \\
   2\dint \sec^3\theta d\theta & = & \sec\theta\tan\theta+\ln(\sec\theta+\tan\theta)+C \\
   \dint \sec^3\theta d\theta & = & \dfrac{1}{2}\sec\theta\tan\theta+\dfrac{1}{2}\ln(\sec\theta+\tan\theta)+\dfrac{C}{2}
   \end{array}$

When $x=-a$, $u=a$.

When $x=a$, $u=-a$.

$\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{ , for $u$ is a dummy variable.}\\
= & \dint_{-a}^a g(-x)h(x) dx \text{ , $\because h(x)=h(-x)$ for all $x\in\mathbb{R}$.}
\end{array}$

Since $h(x)=h(-x)$ for all $x\in\mathbb{R}$, then $h(x)$ is an even function. Hence, we have

$\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{ , $g(x)+g(-x)=1$ for all $x\in\mathbb{R}$.} \\
= & \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{ , proved above.} \\
= & \dfrac{1}{2} \times 2 \dint_{-a}^a g(x)h(x) dx \\
= & \dint_{-a}^a g(x)h(x) dx
\end{array}$

Checking:

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

Also,

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

Hence, we can apply the result of (b) in part (c). For $a=1$, we have

$\begin{array}{cl}
& \dint_{-1}^1 \dfrac{3^x x^2}{(3^x+3^{-x})\sqrt{x^2+1}}dx \\
= & \dint_{-1}^1 g(x)h(x) dx \\
= & \dint_0^1 h(x)dx \text{ , by the result of (b).} \\
= & \dint_0^1\dfrac{x^2}{\sqrt{x^2+1}}dx
\end{array}$

Let $x=\tan\theta$. Then $dx =\sec^2\theta d\theta$.

When $x=0$, $\theta=0$.

When $x=1$, $\theta=\dfrac{\pi}{4}$.

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