3.8 解答

$\begin{displaymath}\begin{array}{llll} x&0 &\Rightarrow & \frac{\pi}{2}\ \hline t&1 &\Rightarrow & 0 \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\cos^{4}{x}\sin{x}} dx$	$\displaystyle =$	$\displaystyle \int_{1}^{0}{t^{4}(-dt)}$
	$\displaystyle =$	$\displaystyle \int_{0}^{1}t^{4} dt = \frac{t^{5}}{5}\mid_{0}^{1} = \frac{1}{5}$

(b) $t = \sqrt{1 + \cos{x}}$ とおくと $t^2 = 1 + \cos{x}$ より $2tdt = -\sin{x}dx$ ．このとき

$\begin{displaymath}\begin{array}{llll} x&0 &\Rightarrow & \frac{\pi}{2}\ \hline t&\sqrt{2} &\Rightarrow & 1 \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{\sqrt{1 + \cos{x}}}} dx$	$\displaystyle =$	$\displaystyle \int_{\sqrt{2}}^{1}{\frac{-2tdt}{t}}$
	$\displaystyle =$	$\displaystyle -2\int_{0}^{\frac{\pi}{2}} dt = -2t\mid_{\sqrt{2}}^{1} = -2(1 - \sqrt{2}) = 2(\sqrt{2} - 1)$

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{4}{x}} dx$

$\displaystyle =$

$\displaystyle \frac{3!!}{4!!}\frac{\pi}{2} = \frac{3\pi}{16}$

$\displaystyle \int{f(x)g'(x)} dx = f(x)g(x) - \int{g(x)f'(x)} dx$

$\begin{displaymath}\begin{array}{ll} f(x) = x^2 & g'(x) = \cos{x}\\ f'(x) = 2x & g(x) = \sin{x} \end{array}\end{displaymath}$

$\displaystyle \int_{-1}^{1}{x^{2}\cos{x}} dx$	$\displaystyle =$	$\displaystyle x^{2}\sin{x}\mid_{-1}^{1} - \int_{-1}^{1}{2x\sin{x}} dx$
	$\displaystyle =$	$\displaystyle \sin{1} - \sin{(-1)} - \int_{-1}^{1}{2x\sin{x}} dx = 2\sin{1} - \int_{-1}^{1}{2x\sin{x}} dx$

$\begin{displaymath}\begin{array}{ll} f(x) = 2x & g'(x) = \sin{x}\\ f'(x) = 2 & g(x) = -\cos{x} \end{array}\end{displaymath}$

$\displaystyle \int_{-1}^{1}{2x\sin{x}} dx$	$\displaystyle =$	$\displaystyle -2x\cos{x}\mid_{-1}^{1} + 2\int_{-1}^{1}{\cos{x}} dx$
	$\displaystyle =$	$\displaystyle -2\cos{1} - 2\cos{(-1)} + 2\sin{x}\mid_{-1}^{1}$
	$\displaystyle =$	$\displaystyle -2\cos{1} - 2\cos{1} + 2\sin{1} - 2\sin{(-1)} = -4\cos{1} + 4\sin{1}$

$\displaystyle \int_{-1}^{1}{x^{2}\cos{x}} dx$	$\displaystyle =$	$\displaystyle 2\sin{1} -[-4\cos{1} + 4\sin{1}]$
	$\displaystyle =$	$\displaystyle 4\cos{1} - 2\sin{1}$

$\displaystyle \int_{0}^{\pi}\cos{nx} dx$

$\displaystyle =$

$\displaystyle \frac{1}{n}\sin{nx}\mid_{0}^{\pi} = 0$

(f) $x = 2\cos{\theta}$ とおくと， $dx = -2\sin{\theta}d\theta$ , $\sqrt{4 -x^{2}} = 2\sin{\theta}$ ．ここで， $\begin{displaymath}\begin{array}{c\vert lcl} x & 0 \rightarrow & 1\ \hline \theta & \frac{\pi}{2} & \rightarrow & \frac{\pi}{3} \end{array}\end{displaymath}$ より，

$\displaystyle \int_{0}^{1}\frac{x^{2}}{\sqrt{4-x^{2}}} dx$	$\displaystyle =$	$\displaystyle -4\int_{\pi/2}^{\pi/3}\frac{4\cos^{3}{\theta}(-2\sin{\theta})}{2\sin{\theta}}d\theta$
	$\displaystyle =$	$\displaystyle -4\int_{\pi/2}^{\pi/3}\cos^{2}{\theta}d\theta = -4\int_{\pi/2}^{\pi/3}\frac{1 + \cos{2\theta}}{2}d\theta$
	$\displaystyle =$	$\displaystyle -2\left[\theta + \frac{\sin{2\theta}}{2}\right]_{\pi/2}^{\pi/3}$
	$\displaystyle =$	$\displaystyle -2\left(\frac{\pi}{3} + \frac{\sin{\frac{2\pi}{3}}}{2} - (-\frac{\pi}{2} + \frac{\sin{\pi}}{2})\right)$
	$\displaystyle =$	$\displaystyle \frac{\pi}{3} - \frac{\sqrt{3}}{2}$

$\displaystyle \int{udv} = uv - \int{vdu}$

$\begin{displaymath}\begin{array}{ll} u = x & dv = e^{x}dx\\ du = dx & v = e^{x} \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{1}xe^{x} dx$	$\displaystyle =$	$\displaystyle xe^{x}\mid_{0}^{1} - \int_{0}^{1}e^{x}dx$
	$\displaystyle =$	$\displaystyle e - e^{x}\mid_{0}^{1} = e - (e - 1) = 1$

2. $\displaystyle{I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{n}{x}dx = \int_{0}^{\frac{\pi}{2}}\sin^{n-1}{x} \sin{x}dx = \frac{n-1}{n} I_{n-2}}$

$\displaystyle{J_{n} = \int_{0}^{\frac{\pi}{2}}\cos^{n}{x}dx = \int_{0}^{\frac{\pi}{2}}\cos^{n-1}{x} \cos{x}dx = \frac{n-1}{n} J_{n-2}}$

$\displaystyle F(-x) = \int_{x}^{-x}f(t)dt = -\int_{-x}^{x}f(t)dt = -F(x)$

$\displaystyle f'(-x)$	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{f(-x+h) - f(-x)}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{f(-(x-h)) - f(x)}{h}$
	$\displaystyle =$	$\displaystyle \lim_{h \to 0}\frac{-(f(x) - f(x-h)}{h} = -f'(x)$

(d) $f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}$ とおくと， $g(x) = \frac{f(x) + f(-x)}{2}$ は偶関数． $h(x) = \frac{f(x) - f(-x)}{2}$ は奇関数となる．なぜならば，

$\displaystyle g(-x) = \frac{f(-x) + f(x)}{2} = g(x), h(-x) = \frac{f(-x) - f(x)}{2} = -h(x)$

4. $x = \tan{t}$ とおくと， $x^2 + 1 = \sec^{2}{t}$ , $dx = \sec^{2}{t}dt$ , $\begin{displaymath}\begin{array}{c\vert ccc} x & 0 & \to 1\\ t & 0 & \to \pi/4 \end{array}\end{displaymath}$ . よって，

$\displaystyle \int_{0}^{\pi/4}\log(\tan{t} + 1)\;dt$

$\displaystyle I$	$\displaystyle =$	$\displaystyle \int_{0}^{\pi/4}\log(\tan{t} + 1)\;dt = \int_{0}^{\pi/4}\log(\frac{2}{\tan{u} + 1})\;du$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi/4}\log{2}du - I = \frac{\pi \log{2}}{4} - I$