3.8 解答

3.8

1.

(a) $t = \cos{x}$とおくと $dt = -\sin{x}dx$．このとき

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

に注意すると

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

$$= \int_{0}^{1}t^{4} dt = \frac{t^{5}}{5}\mid_{0}^{1} = \frac{1}{5}$$

(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

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

に注意すると

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

$$= -2\int_{0}^{\frac{\pi}{2}} dt = -2t\mid_{\sqrt{2}}^{1} = -2(1 - \sqrt{2}) = 2(\sqrt{2} - 1)$$

(c)

$$\int_{0}^{\frac{\pi}{2}}{\sin^{4}{x}} dx = \frac{3!!}{4!!}\frac{\pi}{2} = \frac{3\pi}{16}$$

(d) 部分積分の復習

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

より

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

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

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

ここで，もう一度部分積分を用いると

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

より

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

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

$$= -2\cos{1} - 2\cos{1} + 2\sin{1} - 2\sin{(-1)} = -4\cos{1} + 4\sin{1}$$

したがって，

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

$$= 4\cos{1} - 2\sin{1}$$

(e) $n > 0$のとき

$$\int_{0}^{\pi}\cos{nx} dx = \frac{1}{n}\sin{nx}\mid_{0}^{\pi} = 0$$

(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， $$\begin{array}{c\vert lcl} x & 0 \rightarrow & 1\ \hline \theta & \frac{\pi}{2} & \rightarrow & \frac{\pi}{3} \end{array}$$より，

$$\int_{0}^{1}\frac{x^{2}}{\sqrt{4-x^{2}}} dx = -4\int_{\pi/2}^{\pi/3}\frac{4\cos^{3}{\theta}(-2\sin{\theta})}{2\sin{\theta}}d\theta$$

$$= -4\int_{\pi/2}^{\pi/3}\cos^{2}{\theta}d\theta = -4\int_{\pi/2}^{\pi/3}\frac{1 + \cos{2\theta}}{2}d\theta$$

$$= -2\left[\theta + \frac{\sin{2\theta}}{2}\right]_{\pi/2}^{\pi/3}$$

$$= -2\left(\frac{\pi}{3} + \frac{\sin{\frac{2\pi}{3}}}{2} - (-\frac{\pi}{2} + \frac{\sin{\pi}}{2})\right)$$

$$= \frac{\pi}{3} - \frac{\sqrt{3}}{2}$$

(g)

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

より

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

よって，

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

$$= e - e^{x}\mid_{0}^{1} = e - (e - 1) = 1$$

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

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

3.

(a) $F(-x) = -F(x)$を示せばよい．

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

(b) $f'(-x) = -f'(x)$を示す．

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

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

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

(c) $f'(x) = f(x) - f(-x)(-1)$. ここで，3(a)より，$f(x)$は奇関数．したがって， $f'(x) = f(x) - f(x) = 0$

(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，

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

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

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

次に， $t = \frac{\pi}{4} - u$とおくと， $\tan(\frac{\pi}{4} - u) = \frac{1 - \tan{u}}{1 + \tan{u}}$より，

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

$$= \int_{0}^{\pi/4}\log{2}du - I = \frac{\pi \log{2}}{4} - I$$

したがって，

$$I = \frac{\pi \log{2}}{8}$$