第5章 5.19

(1)

$s = \int_{-\sqrt{2}}^{\sqrt{2}} \sqrt{1+x^2}\,dx = \left[\frac{1}{2}\left(x\sqrt{x^2+1}+\log\left|x+\sqrt{x^2+1}\right|\right)\right]_{-\sqrt{2}}^{\sqrt{2}} = \sqrt{6}+\log(\sqrt{2}+\sqrt{3})$

(2)

$y = \log(1-x^2) \qquad \left[0,\frac{1}{3}\right]$

$s = \int_0^{\frac{1}{3}} \sqrt{1+\left(\frac{2x}{1-x^2}\right)^2}\,dx = \int_0^{\frac{1}{3}} \frac{\sqrt{1-2x^2+x^4+4x^2}}{1-x^2}\,dx = \int_0^{\frac{1}{3}} \frac{\sqrt{1+2x^2+x^4}}{1-x^2}\,dx = \int_0^{\frac{1}{3}} \frac{1+x^2}{1-x^2}\,dx$

$= \int_0^{\frac{1}{3}} \left(-1+\frac{2}{1-x^2}\right)dx = \left[-x-\log\left|\frac{x-1}{x+1}\right|\right]_0^{\frac{1}{3}} = -\frac{1}{3}-\log\frac{1}{2} = -\frac{1}{3}+\log 2$

(3)

$y = \frac{a}{2}\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right) \qquad (0,a)\text{から}(x,y)$

$s = \int_0^x \sqrt{1+\left\{\frac{a}{2}\cdot\frac{1}{a}\left(e^{\frac{x}{a}}-e^{-\frac{x}{a}}\right)\right\}^2}\,dx = \int_0^x \sqrt{1+\frac{1}{4}\left(e^{\frac{x}{a}}-e^{-\frac{x}{a}}\right)^2}\,dx = \int_0^x \sqrt{\frac{1}{4}\left(e^{\frac{2x}{a}}+2+e^{-\frac{2x}{a}}\right)}\,dx$

$= \int_0^x \sqrt{\frac{1}{4}\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)^2}\,dx = \int_0^x \frac{1}{2}\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)dx = \left[\frac{1}{2}a\left(e^{\frac{x}{a}}-e^{-\frac{x}{a}}\right)\right]_0^x = \frac{a}{2}\left(e^{\frac{x}{a}}-e^{-\frac{x}{a}}\right)$

(4)

$y^2 = 8x \qquad (4,4\sqrt{2}) \qquad y=\sqrt{8x}$

$s = \int_0^4 \sqrt{1+\left(\frac{4}{\sqrt{8x}}\right)^2}\,dx = \int_0^4 \sqrt{1+\frac{2}{x}}\,dx = \int_0^4 \sqrt{\frac{x+2}{x}}\,dx$

$\sqrt{\frac{x+2}{x}}=t \ \text{とおくと} \qquad dx=\frac{-4t}{(t^2-1)^2}\,dt \qquad x:0\to4 \qquad t:\infty\to\sqrt{\frac{3}{2}} \quad \text{これから}$

$= \int_\infty^{\sqrt{\frac{3}{2}}} t\cdot\frac{-4t}{(t^2-1)^2}\,dt = \int_{\sqrt{\frac{3}{2}}}^\infty \frac{4t^2}{(t^2-1)^2}\,dt = 2\left\{\left[-\frac{1}{t^2-1}\cdot t\right]_{\sqrt{\frac{3}{2}}}^\infty + \int_{\sqrt{\frac{3}{2}}}^\infty \frac{1}{t^2-1}\,dt\right\} = 2\left\{2\sqrt{\frac{3}{2}}+\left[\frac{1}{2}\log\left|\frac{t-1}{t+1}\right|\right]_{\sqrt{\frac{3}{2}}}^\infty\right\}$

$= 2\left\{2\sqrt{\frac{3}{2}}-\frac{1}{2}\log\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right\} = 2\sqrt{6}+\log\frac{(\sqrt{3}+\sqrt{2})^2}{3-2} = 2\sqrt{6}+2\log(\sqrt{3}+\sqrt{2})$

(5)

$x = a(\cos\theta + \theta\sin\theta) \qquad y = a(\sin\theta - \theta\cos\theta) \qquad 0 \le \theta \le b$

$s = \int_0^b \sqrt{\{a(-\sin\theta+\sin\theta+\theta\cos\theta)\}^2 + \{a(\cos\theta-\cos\theta+\theta\sin\theta)\}^2}\,d\theta$

$= \int_0^b \sqrt{a^2\theta^2(\cos^2\theta+\sin^2\theta)}\,d\theta = \int_0^b a\theta\,d\theta = \left[\frac{1}{2}a\theta^2\right]_0^b = \frac{1}{2}ab^2$