{"id":741,"date":"2025-07-10T19:15:21","date_gmt":"2025-07-10T10:15:21","guid":{"rendered":"https:\/\/math-friend.com\/?p=741"},"modified":"2025-08-01T09:37:37","modified_gmt":"2025-08-01T00:37:37","slug":"%e3%80%90%e4%ba%ac%e9%83%bd%e5%a4%a7%e5%ad%a6%e5%85%a5%e8%a9%a6%e3%80%91%e8%a4%87%e7%b4%a0%e6%95%b0%e3%81%ae%e7%b5%b6%e5%af%be%e5%80%a4%e3%81%ae%e6%9c%80%e5%a4%a7%e6%9c%80%e5%b0%8f%e5%95%8f%e9%a1%8c","status":"publish","type":"post","link":"https:\/\/math-friend.com\/?p=741","title":{"rendered":"\u3010\u4eac\u90fd\u5927\u5b66\u5165\u8a66\u3011\u8907\u7d20\u6570\u306e\u7d76\u5bfe\u5024\u306e\u6700\u5927\u6700\u5c0f\u554f\u984c\u3068\u5c11\u3057\u5de5\u592b\u306e\u3044\u308b\u5b9a\u7a4d\u5206(2025)"},"content":{"rendered":"\n<p>\u4eca\u56de\u306f\u3053\u3061\u3089\u306e\u554f\u984c\u3092\u89e3\u3044\u3066\u3044\u304d\u307e\u3059.<\/p>\n\n\n\n<p class=\"has-border -border02\">\u554f1. \u8907\u7d20\u6570\\(z\\)\u304c\\(|z|=2\\)\u3092\u6e80\u305f\u3059\u8907\u7d20\u6570\u5168\u4f53\u3092\u52d5\u304f\u3068\u304d, \\(\\left|z-\\frac{i}{z}\\right|\\)\u306e\u6700\u5927\u5024\u3068\u6700\u5c0f\u5024\u3092\u6c42\u3081\u3088.<br><br>\u554f2. \u4ee5\u4e0b\u306e\u5b9a\u7a4d\u5206\u306e\u5024\u3092\u6c42\u3081\u3088.<br><br>(1) \\(\\displaystyle \\int_0^{\\sqrt{3}}\\frac{x\\sqrt{x^2+1}+2x^3+1}{x^2+1}dx\\)<br><br>(2) \\(\\displaystyle\\int_0^{\\frac{\\pi}{2}}\\sqrt{\\frac{1-\\cos{x}}{1+\\cos{x}}}dx\\)<br><br><span style=\"text-align:right;display:block;\">(2025 \u4eac\u90fd\u5927\u5b66\u7406\u7cfb[1])<\/span><\/p>\n\n\n\n<p>\u3053\u3061\u3089\u306f\u4eac\u90fd\u5927\u5b66\u306e\u7406\u7cfb\u306e\u5165\u8a66\u554f\u984c\u3067\u3059\u304c, \u5c0f\u554f\u96c6\u3068\u3044\u3046\u3053\u3068\u3067\u96e3\u6613\u5ea6\u306f\u305d\u308c\u307b\u3069\u9ad8\u304f\u3042\u308a\u307e\u305b\u3093. \u6700\u5f8c\u306b\u5225\u89e3\u3082\u7d39\u4ecb\u3057\u307e\u3059.<br><br>\u305d\u308c\u3067\u306f\u89e3\u3044\u3066\u3044\u304d\u307e\u3057\u3087\u3046.<\/p>\n\n\n\n<div class=\"wp-block-group is-style-crease\"><div class=\"wp-block-group__inner-container is-layout-constrained wp-block-group-is-layout-constrained\">\n<p>\u554f1<br>\\(z\\)\u306e\u7d76\u5bfe\u5024\u306f\\(2\\)\u306a\u306e\u3067, \u6975\u5f62\u5f0f\u3067\u8868\u3059\u3068\\(z=2\\left(\\cos{\\theta}+i\\sin{\\theta}\\right)\\)\u3068\u306a\u308a, \\(\\theta\\)\u306f\\(0\\leq \\theta&lt; 2\\pi\\)\u306e\u7bc4\u56f2\u3092\u52d5\u304f\u3068\u8003\u3048\u3066\u826f\u3044. \u3053\u306e\u3068\u304d, \u30c9\u30fb\u30e2\u30a2\u30d6\u30eb\u306e\u5b9a\u7406\u304b\u3089,<br>$$<br>\\begin{align}<br>\\frac{1}{z}&amp;=z^{-1}=\\left\\{2\\left(\\cos{\\theta}+i\\sin{\\theta}\\right)\\right\\}^{-1}\\\\[1.5ex]<br>&amp;=\\frac{1}{2}\\left(\\cos{(-\\theta)}+i\\sin{(-\\theta)}\\right)\\\\[1.5ex]<br>&amp;=\\frac{1}{2}\\left(\\cos{\\theta}-i\\sin{\\theta}\\right)<br>\\end{align}<br>$$\u3068\u306a\u308b. \u3088\u3063\u3066, <br>$$<br>\\begin{align}<br>\\left|z-\\frac{i}{z}\\right|&amp;=\\left|2\\left(\\cos{\\theta}+i\\sin{\\theta}\\right)-i\\cdot\\frac{1}{2}\\left(\\cos{\\theta}-i\\sin{\\theta}\\right)\\right|\\\\[1.5ex]<br>&amp;=\\left|\\left(2\\cos{\\theta}-\\frac{1}{2}\\sin{\\theta}\\right)+i\\left(2\\sin{\\theta}-\\frac{1}{2}\\cos{\\theta}\\right)\\right|\\\\[1.5ex]<br>&amp;=\\sqrt{\\left(2\\cos{\\theta}-\\frac{1}{2}\\sin{\\theta}\\right)^2+\\left(2\\sin{\\theta}-\\frac{1}{2}\\cos{\\theta}\\right)^2}\\\\[1.5ex]<br>&amp;=\\sqrt{4\\cos^2{\\theta}-2\\cos{\\theta}\\sin{\\theta}+\\frac{1}{4}\\sin^2{\\theta}+4\\sin^2{\\theta}-2\\sin{\\theta}\\cos{\\theta}+\\frac{1}{4}\\cos^2{\\theta}}\\\\[1.5ex]<br>&amp;=\\sqrt{4-4\\cos{\\theta}\\sin{\\theta}+\\frac{1}{4}}\\\\[1.5ex]<br>&amp;=\\sqrt{\\frac{17}{4}-2\\cdot 2\\cos{\\theta}\\sin{\\theta}}\\\\[1.5ex]<br>&amp;=\\sqrt{\\frac{17}{4}-2\\sin{2\\theta}}\\\\<br>\\end{align}<br>$$\u3068\u306a\u308b(\u9014\u4e2d\u306e\u5f0f\u5909\u5f62\u3067\u306f, \\(\\sin^2{\\theta}+\\cos^2{\\theta}=1\\)\u3068, \\(\\sin\\)\u306e\u500d\u89d2\u306e\u516c\u5f0f \\(\\sin{2\\theta}=2\\sin{\\theta}\\cos{\\theta}\\)\u3092\u7528\u3044\u305f).<br><br>\\(0\\leq \\theta &lt; 2\\pi\\)\u306e\u3068\u304d, \\(0\\leq 2\\theta &lt; 4\\pi\\)\u3067\u3042\u308a, \\(\\sin{2\\theta}\\)\u306f\\(-1\\leq \\sin{2\\theta}\\leq 1\\)\u3067\u3042\u308b\u304b\u3089, \\(\\sin{2\\theta}=-1\\), \u3064\u307e\u308a\\(\\theta=\\frac{3\\pi}{4}, \\frac{7\\pi}{4}\\)\u306e\u3068\u304d, \u6700\u5927\u5024<br>$$<br>\\sqrt{\\frac{17}{4}+2}=\\sqrt{\\frac{25}{4}}=\\frac{5}{2}<br>$$\u3092\u3068\u308a, \\(\\sin{2\\theta}=1\\), \u3064\u307e\u308a\\(\\theta=\\frac{\\pi}{4}, \\frac{5\\pi}{4}\\)\u306e\u3068\u304d, \u6700\u5c0f\u5024<br>$$<br>\\sqrt{\\frac{17}{4}-2}=\\sqrt{\\frac{9}{4}}=\\frac{3}{2}<br>$$<br>\u3092\u3068\u308b\u3053\u3068\u304c\u308f\u304b\u308b.<\/p>\n\n\n\n<p>\u554f2 (1)<br>\u88ab\u7a4d\u5206\u95a2\u6570\u3092\u5206\u5b50\u306e\u9805\u3054\u3068\u306b\u5206\u89e3\u3059\u308b\u3068,<br>$$<br>\\begin{align}<br>\\int_0^{\\sqrt{3}}\\frac{x\\sqrt{x^2+1}+2x^3+1}{x^2+1}dx &amp;=\\int_0^{\\sqrt{3}}\\frac{x\\sqrt{x^2+1}}{x^2+1}dx\\\\[1.5ex]<br>&amp;+\\int_0^{\\sqrt{3}}\\frac{2x^3}{x^2+1}dx+\\int_0^{\\sqrt{3}}\\frac{1}{x^2+1}dx<br>\\end{align}<br>$$<br>\u3068\u306a\u308b. \u3053\u3053\u30672\u756a\u76ee\u306e\u7a4d\u5206\u306f\u5206\u5b50\u306e\u6b21\u6570\u304c\u5206\u6bcd\u306e\u6b21\u6570\u4ee5\u4e0a\u306b\u306a\u3063\u3066\u3044\u308b\u306e\u3067\u5272\u308a\u7b97\u3092\u3057\u3066\u6b21\u6570\u3092\u4e0b\u3052\u308b.<br>$$<br>\\frac{2x^3}{x^2+1}=\\frac{2x(x^2+1)-2x}{x^2+1}=2x-\\frac{2x}{x^2+1}<br>$$<br>\u3053\u306e\u7d50\u679c\u3092\u7528\u3044\u30662\u756a\u76ee\u306e\u7a4d\u5206\u306e\u88ab\u7a4d\u5206\u95a2\u6570\u3092\u5909\u5f62\u3059\u308b\u3068, \u6c42\u3081\u305f\u3044\u7a4d\u5206\u306f\u4ee5\u4e0b\u306e4\u3064\u306e\u7a4d\u5206\u306b\u5206\u89e3\u3055\u308c\u308b.<br>$$<br>\\begin{align}<br>\\int_0^{\\sqrt{3}}\\frac{x\\sqrt{x^2+1}+2x^3+1}{x^2+1}dx &amp;=\\int_0^{\\sqrt{3}}\\frac{x\\sqrt{x^2+1}}{x^2+1}dx\\\\[1.5ex]<br>&amp;+\\int_0^{\\sqrt{3}}{2x}dx-\\int_0^{\\sqrt{3}}\\frac{2x}{x^2+1}dx+\\int_0^{\\sqrt{3}}\\frac{1}{x^2+1}dx<br>\\end{align}<br>$$<br>\u3053\u308c\u30921\u3064\u305a\u3064\u8a08\u7b97\u3057\u3066\u3044\u304f.<br><br>\u2460 1\u756a\u76ee\u306e\u7a4d\u5206<br>\u5206\u6bcd\u5206\u5b50\u3092\\(\\sqrt{x^2+1}\\)\u3067\u5272\u308b\u3068, \u4ee5\u4e0b\u306e\u5f62\u306b\u306a\u308b.<br>$$<br>\\int_0^{\\sqrt{3}}\\frac{x\\sqrt{x^2+1}}{x^2+1}dx=\\int_0^{\\sqrt{3}}\\frac{x}{\\sqrt{x^2+1}}dx<br>$$ \\(t=x^2+1\\)\u3068\u304a\u304f\u3068, \u7a4d\u5206\u7bc4\u56f2\u306f,<br>$$<br>\\begin{array}{c|ccc}<br>x &amp; 0 &amp; \\rightarrow &amp; \\sqrt{3} \\\\<br>\\hline<br>t &amp; 1 &amp; \\rightarrow &amp; 4 \\\\<br>\\end{array}<br>$$\u3068\u306a\u308a, \\(dt=2xdx\\)\u3088\u308a, \\(xdx=\\frac{1}{2}dt\\)\u3068\u306a\u308b\u306e\u3067, \u7f6e\u63db\u7a4d\u5206\u3092\u5b9f\u884c\u3057\u3066,<br>$$<br>\\begin{align}<br>\\int_0^{\\sqrt{3}}\\frac{x}{\\sqrt{x^2+1}}dx&amp;=\\frac{1}{2}\\int_1^3\\frac{1}{\\sqrt{t}}dt\\\\<br>&amp;=\\frac{1}{2}\\left[2\\sqrt{t}\\right]_1^3=\\sqrt{4}-\\sqrt{1}=1<br>\\end{align}<br>$$\u3068\u306a\u308b.<br><br>\u2461 2\u756a\u76ee\u306e\u7a4d\u5206<br>$$<br>\\int_0^{\\sqrt{3}}2xdx=\\left[x^2\\right]_0^\\sqrt{3}=3<br>$$<br><br>\u2462 3\u756a\u76ee\u306e\u7a4d\u5206<br>\u5206\u5b50\u304c\u5206\u6bcd\u3092\u5fae\u5206\u3057\u305f\u5f62\u306b\u306a\u3063\u3066\u3044\u308b\u306e\u3067, \u4ee5\u4e0b\u306e\u3088\u3046\u306b\u8a08\u7b97\u304c\u3067\u304d\u308b(\u6163\u308c\u3066\u3044\u306a\u3044\u65b9\u306f, \\(t=x^2+1\\)\u3068\u7f6e\u63db\u3057\u3066\u7f6e\u63db\u7a4d\u5206\u3067\u8a08\u7b97\u3057\u3066\u3082\u826f\u3044).<br>$$<br>\\begin{align}<br>\\int_0^{\\sqrt{3}}\\frac{2x}{x^2+1}dx&amp;=\\int_0^{\\sqrt{3}}\\frac{\\left(x^2+1\\right)^{\\prime}}{x^2+1}dx\\\\[1.5ex]<br>&amp;=\\left[\\log{|x^2+1|}\\right]_0^\\sqrt{3}=\\log{4}-\\log{1}=2\\log{2}<br>\\end{align}<br>$$<br><br>\u2463 4\u756a\u76ee\u306e\u7a4d\u5206<br>\\(x=\\tan{\\theta}\\)\u3068\u304a\u304f\u3068, \u7a4d\u5206\u7bc4\u56f2\u306f,<br>$$<br>\\begin{array}{c|ccc}<br>x &amp; 0 &amp; \\rightarrow &amp; \\sqrt{3} \\\\<br>\\hline<br>\\theta &amp; 0 &amp; \\rightarrow &amp; \\frac{\\pi}{3} \\\\<br>\\end{array}<br>$$\u3068\u306a\u308a, \\(dx=\\frac{1}{\\cos^2{\\theta}}d\\theta\\)\u3088\u308a, \u7f6e\u63db\u7a4d\u5206\u3092\u5b9f\u884c\u3057\u3066,<br>$$<br>\\begin{align}<br>\\int_0^{\\sqrt{3}}\\frac{1}{x^2+1}dx&amp;=\\int_0^{\\frac{\\pi}{3}}\\frac{1}{\\tan^2{\\theta}+1}\\frac{1}{\\cos^2{\\theta}}d\\theta\\\\[1.5ex]<br>&amp;=\\int_0^{\\frac{\\pi}{3}}d\\theta=\\frac{\\pi}{3}<br>\\end{align}<br>$$\u3068\u306a\u308b(\\((\\tan^2{\\theta}+1)\\cdot\\cos^2{\\theta}=1\\)\u306b\u6ce8\u610f).<br><br>\u2460\u304b\u3089\u2463\u3088\u308a\u6c42\u3081\u308b\u3079\u304d\u7a4d\u5206\u306e\u5024\u306f,<br>$$<br>\\begin{align}<br>\\int_0^{\\sqrt{3}}\\frac{x\\sqrt{x^2+1}+2x^3+1}{x^2+1}dx &amp;=1+3-2\\log{2}+\\frac{\\pi}{3}=4-2\\log{2}+\\frac{\\pi}{3}<br>\\end{align}<br>$$\u3068\u6c42\u307e\u308b.<br><br>\u554f2 (2)<br>\u4ee5\u4e0b\u306e\u534a\u89d2\u306e\u516c\u5f0f\u304b\u3089, <br>$$<br>\\begin{align}<br>\\cos^2{\\frac{x}{2}}&amp;=\\frac{1+\\cos{x}}{2}\\\\[1.5ex]<br>\\sin^2{\\frac{x}{2}}&amp;=\\frac{1-\\cos{x}}{2}<br>\\end{align}<br>$$\\(1+\\cos{x}\\), \\(1-\\cos{x}\\)\u306f\u4ee5\u4e0b\u306e\u3088\u3046\u306b\u5909\u5f62\u3067\u304d\u308b. <br>$$<br>\\begin{align}<br>1+\\cos{x}&amp;=2\\cos^2{\\frac{x}{2}}\\\\[1.5ex]<br>1-\\cos{x}&amp;=2\\sin^2{\\frac{x}{2}}\\\\<br>\\end{align}<br>$$<br>\u4eca\u8003\u3048\u3066\u3044\u308b\u7a4d\u5206\u7bc4\u56f2\\(0\\leq x\\leq\\frac{\\pi}{2}\\)\u3067\u306f, \\(\\sin{\\frac{x}{2}}\\geq 0\\), \\(\\cos{\\frac{x}{2}}\\geq 0\\)\u3067\u3042\u308b\u304b\u3089, \\(\\sqrt{\\sin^2{\\frac{x}{2}}}=\\sin{\\frac{x}{2}}\\), \\(\\sqrt{\\cos^2{\\frac{x}{2}}}=\\cos{\\frac{x}{2}}\\)\u3068\u306a\u308b\u3053\u3068\u306b\u6ce8\u610f\u3057\u3066, \u88ab\u7a4d\u5206\u95a2\u6570\u3092\u5909\u5f62\u3057\u3066\u3044\u304f\u3068,<br>$$<br>\\begin{align}<br>\\int_0^{\\frac{\\pi}{2}} \\sqrt{\\frac{1 &#8211; \\cos{x}}{1 + \\cos{x}}} dx&amp;= \\int_0^{\\frac{\\pi}{2}} \\sqrt{\\frac{2\\sin^2{\\frac{x}{2}}}{2\\cos^2{\\frac{x}{2}}}} dx\\\\[1.5ex]<br>&amp;= \\int_0^{\\frac{\\pi}{2}} \\sqrt{\\frac{\\sin^2{\\frac{x}{2}}}{\\cos^2{\\frac{x}{2}}}} dx = \\int_0^{\\frac{\\pi}{2}} \\frac{\\sin{\\frac{x}{2}}}{\\cos{\\frac{x}{2}}} dx<br>\\end{align}<br>$$ \u3068\u306a\u308a\u307e\u3059. \u3053\u3053\u3067, <br>$$<br>\\left(\\cos{\\frac{x}{2}}\\right)^{\\prime}=-\\frac{1}{2}\\sin{\\frac{x}{2}}<br>$$\u3088\u308a, <br>$$<br>\\sin{\\frac{x}{2}}=-2\\left(\\cos{\\frac{x}{2}}\\right)^{\\prime}<br>$$\u3067\u3042\u308b\u306e\u3067, \u7a4d\u5206\u306e\u8a08\u7b97\u3092\u7d9a\u3051\u308b\u3068,<br>$$<br>\\begin{align}<br>\\int_0^{\\frac{\\pi}{2}} \\sqrt{\\frac{1 &#8211; \\cos{x}}{1 + \\cos{x}}} dx&amp;=\\int_0^{\\frac{\\pi}{2}} \\frac{\\sin{\\frac{x}{2}}}{\\cos{\\frac{x}{2}}} dx\\\\[1.5ex]<br>&amp;=-2\\int_0^{\\frac{\\pi}{2}} \\frac{\\left(\\cos{\\frac{x}{2}}\\right)^{\\prime}}{\\cos{\\frac{x}{2}}} dx\\\\[1.5ex]<br>&amp;=-2\\left[\\log{\\left|\\cos{\\frac{x}{2}}\\right|}\\right]_0^{\\frac{\\pi}{2}}\\\\[1.5ex]<br>&amp;=-2\\left(\\log{\\frac{1}{\\sqrt{2}}}-\\log{1}\\right)=\\log{2}<br>\\end{align}<br>$$\u3068\u306a\u308b.<\/p>\n<\/div><\/div>\n\n\n\n<p>\u3044\u304f\u3064\u304b\u6ce8\u610f\u70b9\u3067\u3059. \u3053\u306e\u89e3\u7b54\u3067\u3082\u4f55\u5ea6\u304b\u4f7f\u3063\u3066\u3044\u307e\u3059\u304c, \u5206\u5b50\u304c\u5206\u6bcd\u3092\u5fae\u5206\u3057\u305f\u5f62\u306b\u306a\u3063\u3066\u3044\u308b\u5206\u6570\u306e\u95a2\u6570\u3092\u7a4d\u5206\u3059\u308b\u3068\\(\\log{(\u5206\u6bcd)}\\)\u306b\u306a\u308a\u307e\u3059. \u3053\u308c\u304c\u4f7f\u3048\u308b\u3068\u4f59\u8a08\u306a\u7f6e\u63db\u64cd\u4f5c\u304c\u6e1b\u308b\u306e\u3067\u662f\u975e\u4f7f\u3044\u3053\u306a\u3057\u305f\u3044\u3068\u3053\u308d\u3067\u3059.<br>$$<br>\\int\\frac{\\left(f(x)\\right)^\\prime}{f(x)}dx=\\log{f(x)}+C<br>$$<br><br>\u307e\u305f, \\(\\frac{1}{\\sqrt{t}}\\)\u306e\u7a4d\u5206\u306f, \\(\\frac{1}{\\sqrt{t}}=t^{-\\frac{1}{2}}\\)\u3088\u308a, <br>$$<br>\\int{x^{\\alpha}}dx=\\frac{x^{\\alpha + 1}}{\\alpha + 1}+C<br>$$<br>\u3092\u7528\u3044\u3066,<br>$$<br>\\int{t^{-\\frac{1}{2}}}dt=\\frac{t^{-\\frac{1}{2} + 1}}{-\\frac{1}{2}+ 1}+C=2t^{-\\frac{1}{2}}+C<br>$$\u3088\u308a\u6c42\u307e\u308a\u307e\u3059.<br><br>\u6700\u5f8c\u306b\u554f1\u306e\u5225\u89e3\u30922\u3064, \u554f2(2)\u306e\u5225\u89e3\u30921\u3064\u7d39\u4ecb\u3057\u307e\u3059.<\/p>\n\n\n\n<p class=\"has-border -border03\">\u554f1(\u5225\u89e31)<br>\\(|z|=2\\)\u3088\u308a, <br>$$<br>\\left|z-\\frac{1}{z}\\right|=\\left|\\frac{z^2-i}{z}\\right|=\\frac{\\left|z^2-i\\right|}{\\left|z\\right|}=\\frac{1}{2}\\left|z^2-i\\right|<br>$$\u3068\u306a\u308b.<br><br>\u3053\u3053\u3067\\(z=2(\\cos{\\theta}+i\\sin{\\theta}\\)\u3068\u3059\u308b\u3068, \u30c9\u30fb\u30e2\u30a2\u30d6\u30eb\u306e\u5b9a\u7406\u304b\u3089, \\(z^2=4(\\cos{2\\theta}+i\\sin{2\\theta})\\)\u306a\u306e\u3067, \u7d9a\u304d\u3092\u8a08\u7b97\u3059\u308b\u3068,<br>$$<br>\\begin{align}<br>\\frac{1}{2}\\left|z^2-i\\right|&amp;=\\frac{1}{2}\\left|4\\cos{2\\theta}+i\\left(4\\sin{2\\theta}-1\\right)\\right|\\\\[1.5ex]<br>&amp;=\\frac{1}{2}\\sqrt{\\left(4\\cos{2\\theta}\\right)^2+\\left(4\\sin{2\\theta}-1\\right)^2}\\\\[1.5ex]<br>&amp;=\\frac{1}{2}\\sqrt{16\\cos^2{2\\theta}+16\\sin^2{2\\theta}-8\\sin{2\\theta}+1}\\\\[1.5ex]<br>&amp;=\\frac{1}{2}\\sqrt{17-8\\sin{2\\theta}}\\\\<br>\\end{align}<br>$$\u3068\u306a\u308a\u307e\u3059. <br><br>\\(0\\leq \\theta &lt; 2\\pi\\)\u306e\u3068\u304d, \\(0\\leq 2\\theta &lt; 4\\pi\\)\u3067\u3042\u308a, \\(\\sin{2\\theta}\\)\u306f\\(-1\\leq \\sin{2\\theta}\\leq 1\\)\u3067\u3042\u308b\u304b\u3089, \\(\\sin{2\\theta}=-1\\), \u3064\u307e\u308a\\(\\theta=\\frac{3\\pi}{4}, \\frac{7\\pi}{4}\\)\u306e\u3068\u304d, \u6700\u5927\u5024<br>$$<br>\\frac{1}{2}\\sqrt{17+8}=\\frac{5}{2}<br>$$\u3092\u3068\u308a, \\(\\sin{2\\theta}=1\\), \u3064\u307e\u308a\\(\\theta=\\frac{\\pi}{4}, \\frac{5\\pi}{4}\\)\u306e\u3068\u304d, \u6700\u5c0f\u5024<br>$$<br>\\frac{1}{2}\\sqrt{17-8}=\\frac{3}{2}<br>$$<br>\u3092\u3068\u308b\u3053\u3068\u304c\u308f\u304b\u308b.<\/p>\n\n\n\n<p class=\"has-border -border03\">\u554f1(\u5225\u89e32)<br>\\(|z|=2\\)\u3088\u308a, <br>$$<br>\\left|z-\\frac{1}{z}\\right|=\\left|\\frac{z^2-i}{z}\\right|=\\frac{\\left|z^2-i\\right|}{\\left|z\\right|}=\\frac{1}{2}\\left|z^2-i\\right|<br>$$\u3068\u306a\u308b.<br><br>\u3053\u3053\u3067, \\(\\left|z^2-i\\right|\\)\u306f\u8907\u7d20\u6570\u5e73\u9762\u4e0a\u3067\\(z^2\\)\u3068\\(i\\)\u306e\u8ddd\u96e2\u3067\u3042\u308b. \u3053\u3053\u3067\\(z=2(\\cos{\\theta}+i\\sin{\\theta}\\)\u3068\u3059\u308b\u3068, \u30c9\u30fb\u30e2\u30a2\u30d6\u30eb\u306e\u5b9a\u7406\u304b\u3089, \\(z^2=4(\\cos{2\\theta}+i\\sin{2\\theta})\\)\u3068\u306a\u308a, \\(\\left|z^2\\right|\\)\u306f\u8907\u7d20\u6570\u5e73\u9762\u4e0a\u3067\u539f\u70b9(\\0\\)\u3092\u4e2d\u5fc3\u3068\u3059\u308b\u534a\u5f84\\(4\\)\u306e\u5186\u5468\u4e0a\u3092(\u304f\u307e\u306a\u304f)\u52d5\u304f.<br><br>2\u3064\u306e\u5186\u304c\u5185\u63a5\u3059\u308b\u3068\u304d, \u305d\u306e\u63a5\u70b9\u30682\u3064\u306e\u4e2d\u5fc3\u304c1\u76f4\u7dda\u4e0a\u306b\u306a\u308b\u3053\u3068\u3092\u8003\u616e\u3059\u308c\u3070, \u70b9\\(i\\)\u3068\u5186\u5468\u4e0a\u306e\u70b9\u306e\u8ddd\u96e2\u306e\u6700\u5c0f\u5024, \u6700\u5927\u5024\u306f\u305d\u308c\u305e\u308c\\(3\\), \\(5\\)\u3068\u306a\u308b. <br><br><img decoding=\"async\" width=\"728\" height=\"562\" class=\"wp-image-861\" style=\"width: 500px;text-align:center;\" src=\"http:\/\/math-friend.com\/wp-content\/uploads\/2025\/07\/e9731ee377d028ae65ca96ab92bf87b5.png\" alt=\"\" srcset=\"https:\/\/math-friend.com\/wp-content\/uploads\/2025\/07\/e9731ee377d028ae65ca96ab92bf87b5.png 728w, https:\/\/math-friend.com\/wp-content\/uploads\/2025\/07\/e9731ee377d028ae65ca96ab92bf87b5-300x232.png 300w\" sizes=\"(max-width: 728px) 100vw, 728px\" \/><br><br>\u3088\u3063\u3066, \u3053\u308c\u3092\\(\\frac{1}{2}\\)\u500d\u3057\u305f\\(\\frac{3}{2}\\), \\(\\frac{5}{2}\\)\u304c\u305d\u308c\u305e\u308c\u6700\u5c0f\u5024, \u6700\u5927\u5024\u3068\u306a\u308b.<\/p>\n\n\n\n<p>\u3053\u306e\u554f1(\u5225\u89e32)\u306e\u3088\u3046\u306b\u56f3\u5f62\u7684\u306a\u610f\u5473\u3082\u8003\u3048\u306a\u304c\u3089\u554f\u984c\u3092\u89e3\u304f\u3068\u4eca\u5f8c\u89e3\u3051\u308b\u554f\u984c\u306e\u5e45\u304c\u5e83\u304c\u308b\u3067\u3057\u3087\u3046.<br><br>\u6700\u5f8c\u306b\u554f2(2)\u306e\u5225\u89e3\u3067\u3059. \u3053\u308c\u306f\u4e0a\u306e\u89e3\u7b54\u3067\u3042\u3063\u305f\u534a\u89d2\u306e\u516c\u5f0f\u3092\u4f7f\u3063\u305f\u5909\u63db\u304c\u601d\u3044\u3064\u304b\u306a\u3044\u65b9\u5411\u3051\u306e\u89e3\u7b54\u3067\u3059.<\/p>\n\n\n\n<p class=\"has-border -border03\">\u554f2(2)(\u5225\u89e3)<br><br>$$<br>\\int_0^{\\frac{\\pi}{2}}\\sqrt{\\frac{1-\\cos{x}}{1+\\cos{x}}}dx<br>$$<br>\u306b\u3066, \\(t=\\cos{x}\\)\u3068\u304a\u304f\u3068, \u7a4d\u5206\u7bc4\u56f2\u306f,<br>$$<br>\\begin{array}{c|ccc}<br>x &amp; 0 &amp; \\rightarrow &amp; \\frac{\\pi}{2} \\\\<br>\\hline<br>t &amp; 1 &amp; \\rightarrow &amp; 0 \\\\<br>\\end{array}<br>$$\u3068\u306a\u308b. \u307e\u305f, \\(dt=-\\sin{x}dx\\)\u3067\u3042\u308b\u304c, \u4eca\u8003\u3048\u3066\u3044\u308b\u7a4d\u5206\u7bc4\u56f2\\(0\\leq x\\leq \\frac{\\pi}{2}\\)\u3067\u306f, \\(\\sin{x}\\geq 0\\)\u3088\u308a, <br>$$<br>\\sin{x}=\\sqrt{1-\\cos^2{x}}<br>$$<br>\u3067\u3042\u308b, \u3053\u3053\u3067, \\(\\cos{x}=t\\)\u3088\u308a, <br>$$<br>dt=-\\sin{x}dx=-\\sqrt{1-\\cos^2{x}}dx=-\\sqrt{1-t^2}dx<br>$$\u3067\u3042\u308b\u304b\u3089, <br>$$<br>dx=-\\frac{1}{\\sqrt{1-t^2}}dt<br>$$\u3068\u306a\u308a, \\(dx\\)\u3092\\(dt\\)\u306b\u5909\u63db\u3067\u304d\u308b.<br><br>\u4ee5\u4e0a\u304b\u3089, \u7f6e\u63db\u7a4d\u5206\u3092\u884c\u3046\u3068,<br>$$<br>\\begin{align}<br>\\int_0^{\\frac{\\pi}{2}}\\sqrt{\\frac{1-\\cos{x}}{1+\\cos{x}}}dx&amp;=\\int_1^0\\sqrt{\\frac{1-t}{1+t}}\\left(-\\frac{1}{\\sqrt{1-t^2}}\\right)dt\\\\[1.5ex]<br>&amp;=\\int_0^1\\frac{\\sqrt{1-t}}{\\sqrt{{1+t}}}\\frac{1}{\\sqrt{1-t}\\sqrt{1+t}}dt\\\\[1.5ex]<br>&amp;=\\int_0^1\\frac{1}{1+t}dt\\\\<br>\\end{align}<br>$$\u3067\u3042\u308b(\u9014\u4e2d\u3067, \\(-(\u30de\u30a4\u30ca\u30b9)\\)\u3092\u4f7f\u3063\u3066\u7a4d\u5206\u533a\u9593\u306e\u4e0a\u7aef\u3068\u4e0b\u7aef\u3092\u4ea4\u63db\u3057\u3066\u3044\u308b)\u3053\u3068\u306b\u6ce8\u610f.<br><br>\u3053\u306e\u7a4d\u5206\u306f\u5206\u5b50\u304c\u5206\u6bcd\u3092\u5fae\u5206\u3057\u305f\u5f62\u306b\u306a\u3063\u3066\u3044\u308b\u306e\u3067,<br><br>$$<br>\\int_0^1\\frac{1}{1+t}dt=\\left[\\log{1+t}\\right]_0^1=\\log{2}<br>$$\u3068\u306a\u308b.<\/p>\n\n\n\n<p>youtube\u3067\u3082\u89e3\u8aac\u3057\u3066\u3044\u307e\u3059.<\/p>\n\n\n\n<iframe width=\"560\" height=\"315\" src=\"https:\/\/www.youtube.com\/embed\/aSWneBxUo-U?si=g4WI5nsk6IzxNcDv\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe>\n\n\n\n\n","protected":false},"excerpt":{"rendered":"<p>\u4eca\u56de\u306f\u3053\u3061\u3089\u306e\u554f\u984c\u3092\u89e3\u3044\u3066\u3044\u304d\u307e\u3059. \u554f1. \u8907\u7d20\u6570\\(z\\)\u304c\\(|z|=2\\)\u3092\u6e80\u305f\u3059\u8907\u7d20\u6570\u5168\u4f53\u3092\u52d5\u304f\u3068\u304d, \\(\\left|z-\\frac{i}{z}\\right|\\)\u306e\u6700\u5927\u5024\u3068\u6700\u5c0f\u5024\u3092\u6c42\u3081\u3088. \u554f2. \u4ee5\u4e0b\u306e\u5b9a\u7a4d [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":742,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"swell_btn_cv_data":"","_themeisle_gutenberg_block_has_review":false,"footnotes":""},"categories":[6],"tags":[],"class_list":["post-741","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-tokimakuru"],"_links":{"self":[{"href":"https:\/\/math-friend.com\/index.php?rest_route=\/wp\/v2\/posts\/741","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math-friend.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/math-friend.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/math-friend.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/math-friend.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=741"}],"version-history":[{"count":144,"href":"https:\/\/math-friend.com\/index.php?rest_route=\/wp\/v2\/posts\/741\/revisions"}],"predecessor-version":[{"id":2182,"href":"https:\/\/math-friend.com\/index.php?rest_route=\/wp\/v2\/posts\/741\/revisions\/2182"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/math-friend.com\/index.php?rest_route=\/wp\/v2\/media\/742"}],"wp:attachment":[{"href":"https:\/\/math-friend.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=741"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/math-friend.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=741"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math-friend.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=741"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}