(2)

(1) において $f(t),$ $g(t)$ をそれぞれ $t,$ $f(t)$ に置き換えると, 求める等式が得られる.

(3)

$x = r(\theta )\cos\theta,$ $y = r(\theta )\sin\theta$ とおく. このとき, \[\begin{aligned} \frac{dx}{d\theta} &= r'(\theta )\cos\theta -r(\theta )\sin\theta, \\ \quad \frac{dy}{d\theta} &= r'(\theta )\sin\theta +r(\theta )\cos\theta \end{aligned}\] となり, \[\begin{aligned} &\left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2 \\ &= \{ r'(\theta )\cos\theta -r(\theta )\sin\theta\} ^2+\{ r'(\theta )\sin\theta +r(\theta )\cos\theta\} ^2 \\ &= r'(\theta )^2(\cos ^2\theta +\sin ^2\theta )+r(\theta )^2(\cos ^2\theta +\sin ^2\theta ) \\ &= r'(\theta )^2+r(\theta )^2 \end{aligned}\] となるから, \[ L = \int _a^b\sqrt{r'(\theta )^2+r(\theta )^2}\,d\theta\] が成り立つ.

　$x = \theta -\sin\theta,$ $y = 1-\cos\theta$ を $\theta$ で微分すると \[\frac{dx}{d\theta} = 1-\cos\theta, \quad \frac{dy}{d\theta} = \sin\theta\] となるから \[\begin{aligned} \left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2 &= (1-\cos\theta )^2+\sin ^2\theta \\ &= 1-2\cos\theta +\cos ^2\theta +\sin ^2\theta \\ &= 2-2\cos\theta = 4\sin ^2\frac{\theta}{2} \\ \end{aligned}\] であり, $0 \leqq \theta \leqq 2\pi$ において $\sin\dfrac{\theta}{2} \geqq 0$ である. ゆえに, 求める長さは, \[\begin{aligned} L &= \int_0^{2\pi}\sqrt{\left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2 }\,d\theta \\ &= \int_0^{2\pi}\sqrt{4\sin ^2\frac{\theta}{2}}\,d\theta \\ &= 2\int_0^{2\pi}\sin\frac{\theta}{2}\,d\theta \\ &= 2\left[ -2\cos\frac{\theta}{2}\right] _0^{2\pi} \\ &= 2\cdot (-2)(-1-1) = 8 \end{aligned}\] である.

　$x = (1+\cos\theta )\cos\theta,$ $y = (1+\cos\theta )\sin\theta$ を $\theta$ で微分すると \[\begin{aligned} \frac{dx}{d\theta} &= (-\sin\theta )\cos\theta +(1+\cos\theta )(-\sin\theta ) \\ &= -2\sin\theta\cos\theta -\sin\theta \\ &= -\sin 2\theta -\sin\theta, \\ \frac{dy}{d\theta} &= (-\sin\theta )\sin\theta +(1+\cos\theta )\cos\theta \\ &= (\cos ^2\theta -\sin ^2\theta )+\cos\theta \\ &= \cos 2\theta +\cos\theta \end{aligned}\] となるから \[\begin{aligned} \left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2 &= (-\sin 2\theta -\sin\theta )^2+(\cos 2\theta +\cos\theta )^2 \\ &= \sin ^22\theta +2\sin 2\theta\sin\theta +\sin ^2\theta \\ &\qquad +\cos ^22\theta +2\cos 2\theta\cos\theta +\cos ^2\theta \\ &= 2+2\cos (2\theta -\theta ) \\ &= 2(1+\cos\theta ) \\ &= 4\cos ^2\frac{\theta}{2} \end{aligned}\] であり, $-\pi \leqq \theta \leqq \pi$ において $\cos\dfrac{\theta}{2} \geqq 0$ である. ゆえに, 求める長さは, \[\begin{aligned} L &= \int_0^{2\pi}\sqrt{\left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2}\,d\theta \\ &= \int_{-\pi}^\pi\sqrt{4\cos ^2\frac{\theta}{2}}\,d\theta \\ &= 2\int_{-\pi}^\pi\cos\frac{\theta}{2}\,d\theta \\ &= 4\int_0^\pi\cos\frac{\theta}{2}\,d\theta \\ &= 4\left[ 2\sin\frac{\theta}{2}\right] _0^\pi \\ &= 4\cdot 2(1-0) = 8 \end{aligned}\] である.

　この曲線は極方程式 \[ r = 1+\cos\theta \quad (-\pi \leqq \theta \leqq \pi )\] で表されるから, 求める長さは, \[\begin{aligned} L &= \int_{-\pi}^\pi\sqrt{r^2+\left(\frac{dr}{d\theta}\right) ^2}\,d\theta \\ &= \int_{-\pi}^\pi\sqrt{(1+\cos\theta )^2+(-\sin\theta )^2}\,d\theta \\ &= \int_{-\pi}^\pi\sqrt{1+2\cos\theta +\cos ^2\theta +\sin ^2\theta}\,d\theta \\ &= \int_{-\pi}^\pi\sqrt{2(1+\cos\theta )}\,d\theta \\ &= \int_{-\pi}^\pi\sqrt{4\cos ^2\frac{\theta}{2}}\,d\theta \\ &= 2\int_{-\pi}^\pi\cos\frac{\theta}{2}\,d\theta \\ &= 4\int_0^\pi\cos\frac{\theta}{2}\,d\theta \\ &= 4\left[ 2\sin\frac{\theta}{2}\right] _0^\pi \\ &= 4\cdot 2(1-0) = 8 \end{aligned}\] である.

\[\cos ^3(2\pi -\theta ) = \cos ^3\theta, \quad \sin ^3(2\pi -\theta ) = -\sin ^3\theta\] から曲線は $x$ 軸に関して対称, \[\cos ^3(\pi -\theta ) = -\cos ^3\theta, \quad \sin ^3(\pi -\theta ) = \sin ^3\theta\] から曲線は $y$ 軸に関して対称である. よって, \[ L = 4\int_0^{\frac{\pi}{2}}\sqrt{\left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2}\,d\theta\] である. $x = \cos ^3\theta,$ $y = \sin ^3\theta$ を $\theta$ で微分すると \[\frac{dx}{d\theta} = -3\cos ^2\theta\sin\theta, \quad \frac{dy}{d\theta} = 3\sin ^2\theta\cos\theta\] となるから \[\begin{aligned} \left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2 &= (-3\cos ^2\theta\sin\theta )^2+(3\sin ^2\theta\cos\theta )^2 \\ &= 9\cos ^4\theta\sin ^2\theta +9\sin ^4\theta\cos ^2\theta \\ &= 9\sin ^2\theta\cos ^2\theta (\cos ^2\theta +\sin ^2\theta ) \\ &= \frac{9}{4}(2\sin\theta\cos\theta )^2 \\ &= \frac{9}{4}\sin ^22\theta \end{aligned}\] であり, $0 \leqq \theta \leqq \dfrac{\pi}{2}$ において $\sin 2\theta \geqq 0$ である. ゆえに, 求める長さは, \[\begin{aligned} L &= 4\int_0^{\frac{\pi}{2}}\sqrt{\frac{9}{4}\sin ^22\theta}\,d\theta \\ &= 4\cdot\frac{3}{2}\int_0^{\frac{\pi}{2}}\sin 2\theta\,d\theta \\ &= 6\left[ -\frac{\cos 2\theta}{2}\right] _0^{\frac{\pi}{2}} \\ &= -3(-1-1) = 6 \end{aligned}\] である.

(1)

$\angle\mathrm{OBQ} = \theta$ のとき $\mathrm Q(\sin\theta,1-\cos\theta )$ であるから, \[\begin{aligned} \overrightarrow{\mathrm{BQ}} &= \overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OB}} \\ &= (\sin\theta,1-\cos\theta )-(0,1) \\ &= (\sin\theta,-\cos\theta ) \end{aligned}\] で, この単位法線ベクトルは \[\pm (\cos\theta,\sin\theta )\] である. $\overrightarrow{\mathrm{QP}}$ は $\vec n = (\cos\theta,\sin\theta )$ を $\mathrm{PQ} = \mathrm{OA}-\stackrel{\frown}{\mathrm{OQ}} = 2\pi -\theta$ 倍に伸ばしたベクトルであるから, \[\begin{aligned} &\overrightarrow{\mathrm{OP}} = \overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{QP}} \\ &= \overrightarrow{\mathrm{OQ}}+(2\pi -\theta )\vec n \\ &= (\sin\theta,1-\cos\theta )+(2\pi -\theta )(\cos\theta,\sin\theta ) \\ &= (\sin\theta +(2\pi -\theta )\cos\theta,1-\cos\theta +(2\pi -\theta )\sin\theta ) \end{aligned}\] である.

(2)

\[\begin{aligned} x &= \sin\theta +(2\pi -\theta )\cos\theta, \\ y &= 1-\cos\theta +(2\pi -\theta )\sin\theta \end{aligned}\] とおくと, \[\begin{aligned} \frac{dx}{d\theta} &= \cos\theta -\cos\theta +(2\pi -\theta )(-\sin\theta ) \\ &= -(2\pi -\theta )\sin\theta, \\ \frac{dy}{d\theta} &= \sin\theta -\sin\theta +(2\pi -\theta )\cos\theta \\ &= (2\pi -\theta )\cos\theta \end{aligned}\] となる. よって, 求める曲線の長さは, \[\begin{aligned} L &= \int _0^{2\pi}\sqrt{\left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2}\,d\theta \\ &= \int _0^{2\pi}(2\pi -\theta )\,d\theta \\ &= \left[ 2\pi\theta -\frac{\theta ^2}{2}\right] _0^{2\pi} \\ &= 2\pi ^2 \end{aligned}\] である.

(A)

(1)

$\{ \log u(x)\}' = \dfrac{u'(x)}{u(x)}$ から, \[ f'(x) = \frac{1+\dfrac{2x}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}} = \frac{\dfrac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}} = \frac{1}{\sqrt{1+x^2}}\] である.

(2)

曲線 $r = \theta$ $(0 \leqq \theta \leqq 2\pi )$ について, $x = r\cos\theta,$ $y = r\sin\theta$ とおく. このとき, \[\begin{aligned} &\frac{dx}{d\theta} = \cos\theta -\theta\sin\theta, \\ &\frac{dy}{d\theta} = \sin\theta +\theta\cos\theta, \\ &\left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2 = 1+\theta ^2 \end{aligned}\] から \[\begin{aligned} L &= \int_0^{2\pi}\sqrt{1+\theta ^2}\,d\theta \\ &= [\theta\sqrt{1+\theta ^2}]_0^{2\pi}-\int_0^{2\pi}\frac{\theta ^2}{\sqrt{1+\theta ^2}}\,d\theta \\ &= 2\pi\sqrt{1+4\pi ^2}-\int_0^{2\pi}\frac{1+\theta ^2-1}{\sqrt{1+\theta ^2}}\,d\theta \\ &= 2\pi\sqrt{1+4\pi ^2}-L+\int_0^{2\pi}\frac{d\theta}{\sqrt{1+\theta ^2}}, \\ 2L &= 2\pi\sqrt{1+4\pi ^2}+[\log (\theta +\sqrt{1+\theta ^2})]_0^{2\pi} \\ &= 2\pi\sqrt{1+4\pi ^2}+\log (2\pi +\sqrt{1+4\pi ^2}) \end{aligned}\] であるので, 求める長さは \[ L = \pi\sqrt{1+4\pi ^2}+\frac{1}{2}\log (2\pi +\sqrt{1+4\pi ^2})\] である.

(B)

曲線 $r = e^\theta$ $(0 \leqq \theta \leqq 2\pi )$ について, $x = r\cos\theta,$ $y = r\sin\theta$ とおく. このとき, \[\begin{aligned} &\frac{dx}{d\theta} = e^\theta (\cos\theta -\sin\theta ), \\ &\frac{dy}{d\theta} = e^\theta (\sin\theta +\cos\theta ), \\ &\left(\frac{dx}{d\theta}\right) ^2+\left(\frac{dy}{d\theta}\right) ^2 \\ &= e^{2\theta}(\cos\theta -\sin\theta )^2+e^{2\theta}(\sin\theta +\cos\theta )^2 \\ &= 2e^{2\theta} \end{aligned}\] となるので, 求める長さは \[ L = \int_0^{2\pi}\sqrt 2e^\theta\,d\theta = \sqrt 2[e^\theta ]_0^{2\pi} = \sqrt 2(e^{2\pi}-1)\] である.

(A)

(2)

求める曲線の長さは \[ L = \int_0^{2\pi}\sqrt{\left(\frac{dr}{d\theta}\right) ^2+r^2}\,d\theta = \int_0^{2\pi}\sqrt{1+\theta ^2}\,d\theta\] で, 以下同様である.

(B)

求める曲線の長さは, \[\begin{aligned} L &= \int_0^{2\pi}\sqrt{\left(\frac{dr}{d\theta}\right) ^2+r^2}\,d\theta = \int_0^{2\pi}\sqrt{e^{2\theta}+e^{2\theta}}\,d\theta \\ &= \sqrt 2\int_0^{2\pi}e^\theta\,d\theta = \sqrt 2[e^\theta ]_0^{2\pi} = \sqrt 2(e^{2\pi}-1) \end{aligned}\] である.

(1)

$f'(x) = \dfrac{e^x-e^{-x}}{2},$ $f''(x) = \dfrac{e^x+e^{-x}}{2} = f(x) \geqq 0$ から \[\begin{aligned} 1+f'(x)^2 &= 1\!+\!\left(\frac{e^x\!-\!e^{-x}}{2}\right) ^2 \!= 1\!+\!\frac{e^{2x}\!-\!2\!+\!e^{-2x}}{4} \\ &= \frac{e^{2x}+2+e^{-2x}}{4} = \left(\frac{e^x+e^{-x}}{2}\right) ^2 \\ &= f(x)^2 = f''(x)^2 \end{aligned}\] であるので, \[\begin{aligned} \overset{\frown}{\mathrm{AP}} &= \int_0^t\sqrt{1+f'(x)^2}\,dx = \int_0^tf''(x)\,dx \\ &= \left[ f'(x)\right] _0^t = f'(t) \end{aligned}\] が成り立つ.

(2)

直線 $\mathrm{QH}$ は点 $(t,0)$ を通り, 接線 $\mathrm{PQ}$ に垂直であるから, その方程式は

$y = -\dfrac{1}{f'(t)}(x-t)$　つまり　$x+f'(t)y-t = 0$

である. よって, 点 $\mathrm P(t,f(t))$ と直線 $\mathrm{QH}$ の距離について \[\begin{aligned} \mathrm{PQ} &= \frac{|t+f'(t)f(t)-t|}{\sqrt{1+f'(t)^2}} = \frac{f'(t)f(t)}{f(t)} \\ &= f'(t) = \overset{\frown}{\mathrm{AP}} \quad (\because (1)) \end{aligned}\] が成り立つ.

(1)

$t = x+\sqrt{x^2+a^2}$ とおくと, \[\frac{dt}{dx} = 1+\frac{2x}{2\sqrt{x^2+a^2}} = \frac{t}{\sqrt{x^2+a^2}}\] よって \[\frac{1}{\sqrt{x^2+a^2}}\cdot\frac{dx}{dt} = \frac{1}{t}\] となるから, 求める不定積分は \[\begin{aligned} &\int\frac{dx}{\sqrt{x^2+a^2}} = \int\frac{dt}{t} \\ &= \log t+C = \log (x+\sqrt{x^2+a^2})+C \end{aligned}\] ($C$: 積分定数) である.

(2)

\[\begin{aligned} &\int\sqrt{x^2+a^2}\,dx = \int (x)'\sqrt{x^2+a^2}\,dx \\ &= x\sqrt{x^2+a^2}-\int x\cdot\frac{2x}{2\sqrt{x^2+a^2}}\,dx \\ &= x\sqrt{x^2+a^2}-\int\frac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}}\,dx \\ &= x\sqrt{x^2+a^2}-\int\sqrt{x^2+a^2}\,dx+a^2\int\frac{dx}{\sqrt{x^2+a^2}} \end{aligned}\] と (1) の結果により, \[\begin{aligned} &\int\sqrt{x^2+a^2}\,dx \\ &= \frac{1}{2}\left( x\sqrt{x^2+a^2}+a^2\int\frac{dx}{\sqrt{x^2+a^2}}\right) \\ &= \frac{1}{2}\left\{ x\sqrt{x^2+a^2}+a^2\log (x+\sqrt{x^2+a^2})\right\} +C \end{aligned}\] ($C$: 積分定数) である.

(3)

(2) の結果から, \[\begin{aligned} L &= \int_0^1\sqrt{1+\left(\frac{dy}{dx}\right) ^2}\,dx \\ &= \int_0^1\sqrt{1+(2x)^2}\,dx = 2\int_0^1\sqrt{x^2+\frac{1}{4}}\,dx \\ &= 2\cdot\frac{1}{2}\left[ x\sqrt{x^2+\frac{1}{4}}+\frac{1}{4}\log\left( x+\sqrt{x^2+\frac{1}{4}}\right)\right] _0^1 \\ &= \frac{\sqrt 5}{2}+\frac{1}{4}\log\left( 1+\frac{\sqrt 5}{2}\right) -\frac{1}{4}\log\frac{1}{2} \\ &= \frac{\sqrt 5}{2}+\frac{1}{4}\log\frac{1+\dfrac{\sqrt 5}{2}}{\dfrac{1}{2}} = \frac{\sqrt 5}{2}+\frac{1}{4}\log (2+\sqrt 5) \end{aligned}\] である.

(2)

$t = x+\sqrt{x^2+a^2}$ とおく. このとき, $t-x = \sqrt{x^2+a^2}$ から $(t-x)^2 = x^2+a^2,$ $t^2-2xt = a^2,$ よって \[ x = \frac{1}{2}\left( t-\frac{a^2}{t}\right)\] となる. また, \[\frac{dx}{dt} = \frac{1}{2}\left( 1+\frac{a^2}{t^2}\right)\] であり, \[\begin{aligned} x^2+a^2 &= \frac{1}{4}\left( t-\dfrac{a^2}{t}\right) ^2+a^2 \\ &= \frac{1}{4}\left( t^2-2a^2+\frac{a^4}{t^2}+4a^2\right) \\ &= \frac{1}{4}\left( t^2+2a^2+\frac{a^4}{t^2}\right) = \frac{1}{4}\left( t+\frac{a^2}{t}\right) ^2, \\ t+\frac{a^2}{t} &= 2\sqrt{x^2+a^2}, \\ t-\frac{a^2}{t} &= 2x \end{aligned}\] であるから, 求める不定積分は \[\begin{aligned} &\int\sqrt{x^2+a^2}\,dx \\ &= \int\sqrt{\frac{1}{4}\left( t+\frac{a^2}{t}\right) ^2}\cdot\frac{1}{2}\left( 1+\frac{a^2}{t^2}\right)\,dt \\ &= \int\frac{1}{2}\left( t+\frac{a^2}{t}\right)\cdot\frac{1}{2}\left( 1+\frac{a^2}{t^2}\right)\,dt \\ &= \frac{1}{4}\int\left( t+\frac{2a^2}{t}+\frac{a^4}{t^3}\right)\,dt \\ &= \frac{1}{4}\left(\frac{t^2}{2}+2a^2\log t-\frac{a^4}{2t^2}\right) +C \\ &= \frac{1}{8}\left( t^2-\frac{a^4}{t^2}\right) +\frac{a^2}{2}\log t+C \\ &= \frac{1}{8}\left( t+\frac{a^2}{t}\right)\left( t-\frac{a^2}{t}\right) +\frac{a^2}{2}\log t+C \\ &= \frac{1}{8}\cdot 2\sqrt{x^2+a^2}\cdot 2x+\frac{a^2}{2}\log (x+\sqrt{x^2+a^2})+C \\ &= \frac{1}{2}\left\{ x\sqrt{x^2+a^2}+a^2\log (x+\sqrt{x^2+a^2})\right\} +C \end{aligned}\] ($C$: 積分定数) である.

(1)

$x = \dfrac{1}{2}\left( t-\dfrac{a^2}{t}\right)$ $(t > 0)$ とおく. このとき, \[\frac{dx}{dt} = \frac{1}{2}\left( 1+\frac{a^2}{t^2}\right)\] であり, \[\begin{aligned} &x^2+a^2 = \frac{1}{4}\left( t-\dfrac{a^2}{t}\right) ^2+a^2 \\ &= \frac{1}{4}\left( t^2-2a^2+\frac{a^4}{t^2}+4a^2\right) = \frac{1}{4}\left( t^2+2a^2+\frac{a^4}{t^2}\right) \\ &= \frac{1}{4}\left( t+\frac{a^2}{t}\right) ^2 = \frac{t^2}{4}\left( 1+\frac{a^2}{t^2}\right) ^2 \end{aligned}\] である. また, $2xt = t^2-a^2$ つまり $t^2-2xt-a^2 = 0$ から \[ t = x+\sqrt{x^2+a^2}\] であるので, 求める不定積分は \[\begin{aligned} \int\frac{dx}{\sqrt{x^2+a^2}} &= \int\frac{\dfrac{1}{2}\left( 1+\dfrac{a^2}{t^2}\right)}{\sqrt{\dfrac{t^2}{4}\left( 1+\dfrac{a^2}{t^2}\right) ^2}}\,dt \\ &= \int\frac{dt}{t} = \log t+C \\ &= \log (x+\sqrt{x^2+a^2})+C \end{aligned}\] ($C$: 積分定数) である.

(1)

$u = \log (x+\sqrt{x^2+a^2})$ とおくと \[\begin{aligned} \frac{du}{dx} &= \frac{1+\dfrac{2x}{2\sqrt{x^2+a^2}}}{x+\sqrt{x^2+a^2}} = \frac{\sqrt{x^2+a^2}+x}{(x+\sqrt{x^2+a^2})\sqrt{x^2+a^2}} \\ &= \frac{1}{\sqrt{x^2+a^2}} \end{aligned}\] から \[\frac{1}{\sqrt{x^2+a^2}}\cdot\frac{dx}{du} = 1\] となるので, 求める不定積分は \[\begin{aligned} \int\frac{dx}{\sqrt{x^2+a^2}} &= \int du = u+C \\ &= \log (x+\sqrt{x^2+a^2})+C \end{aligned}\] ($C$: 積分定数) である.

(1)

$x = \dfrac{e^u-a^2e^{-u}}{2}$ とおく. このとき, \[\frac{dx}{du} = \frac{e^u+a^2e^{-u}}{2}\] であり, \[\begin{aligned} &x^2+a^2 = \left(\frac{e^u-a^2e^{-u}}{2}\right) ^2+a^2 \\ &= \frac{e^{2u}-2a^2+a^4e^{-2u}+4a^2}{4} \\ &= \frac{e^{2u}+2a^2+a^4e^{-2u}}{4} = \frac{(e^u+a^2e^{-u})^2}{4} \end{aligned}\] である. また, $2xe^u = e^{2u}-a^2$ つまり $e^{2u}-2xe^u-a^2 = 0$ から \[ e^u = x+\sqrt{x^2+a^2}, \quad u = \log (x+\sqrt{x^2+a^2})\] であるので, 求める不定積分は \[\begin{aligned} \int\frac{dx}{\sqrt{x^2+a^2}} &= \int\frac{\dfrac{e^u+a^2e^{-u}}{2}}{\sqrt{\dfrac{(e^u+a^2e^{-u})^2}{4}}}\,du \\ &= \int du = u+C \\ &= \log (x+\sqrt{x^2+a^2})+C \end{aligned}\] ($C$: 積分定数) である.