\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\u4e0d\u5b9a\u7a4d\u5206 \\(\\displaystyle\\int e^{-x} \\sin^2 x \\, dx\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\u5b9a\u7a4d\u5206 \\(\\displaystyle\\int _ 0^1 \\sqrt{1 +2 \\sqrt{x}} \\, dx\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u6c42\u3081\u308b\u4e0d\u5b9a\u7a4d\u5206\u3092 \\(I\\) \u3068\u304a\u304f.\r\n\\[\\begin{align}\r\nI & = \\displaystyle\\int e^{-x} \\dfrac{1 -\\cos 2x}{2} \\, dx \\\\\r\n& = -\\dfrac{e^{-x}}{2} -\\dfrac{1}{2} \\underline{\\displaystyle\\int e^{-x} \\cos 2x \\, dx} \\quad ... [1]\r\n\\end{align}\\]\r\n\u3053\u3053\u3067\u4e0b\u7dda\u90e8\u3092 \\(J\\) \u3068\u304a\u304f\u3068, \u90e8\u5206\u7a4d\u5206\u3092\u7528\u3044\u3066\r\n\\[\\begin{align}\r\nJ & = -e^{-x} \\cos 2x +\\dfrac{1}{2} \\displaystyle\\int e^{-x} \\sin 2x \\, dx \\\\\r\n& = e^{-x} \\cos 2x -\\dfrac{e^{-x}}{2} \\sin 2x -\\dfrac{1}{4} \\displaystyle\\int e^{-x} \\cos 2x \\, dx \\\\\r\n& = -\\dfrac{e^{-x}}{2} \\left( 2 \\cos 2x +\\sin 2x \\right) -\\dfrac{J}{4}\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{gather}\r\n\\dfrac{5}{4} J = -\\dfrac{e^{-x}}{2} \\left( 2 \\cos 2x +\\sin 2x \\right) \\\\\r\n\\text{\u2234} \\quad J = -\\dfrac{2 e^{-x}}{5} \\left( 2 \\cos 2x +\\sin 2x \\right)\r\n\\end{gather}\\]\r\n\u3088\u3063\u3066, [1] \u306b\u4ee3\u5165\u3057\u3066\r\n\\[\r\nI = \\underline{-\\dfrac{e^{-x}}{10} \\left( 5 +4 \\cos 2x +2 \\sin 2x \\right) +C}\r\n\\]\r\n\u305f\u3060\u3057, \\(C\\) \u306f\u7a4d\u5206\u5b9a\u6570.<\/p>\r\n (2)<\/strong><\/p>\r\n \u6c42\u3081\u308b\u7a4d\u5206\u5024\u3092 \\(I\\) \u3068\u304a\u304f.
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n
\r\n\\(t = 1 +2 \\sqrt{x}\\) \u3068\u304a\u304f\u3068\r\n\\[\\begin{gather}\r\ndt = \\dfrac{dx}{\\sqrt{x}} \\\\\r\n\\text{\u2234} \\quad dx = \\dfrac{t-1}{2} dt\r\n\\end{gather}\\]\r\n\u307e\u305f\r\n\\[\r\n\\begin{array}{c|ccc} x & 0 & \\rightarrow & 1 \\\\ \\hline t & 1 & \\rightarrow & 3 \\end{array}\r\n\\]\r\n\u306a\u306e\u3067, \u7f6e\u63db\u7a4d\u5206\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nI & = \\displaystyle\\int _ {1}^{3} \\sqrt{t} \\cdot \\dfrac{t-1}{2} \\, dt \\\\\r\n& = \\dfrac{1}{2} \\left[ \\dfrac{2 x^{\\frac{5}{2}}}{5} -\\dfrac{2 x^{\\frac{3}{2}}}{3} \\right] _ {1}^{3} \\\\\r\n& = \\left( \\dfrac{9 \\sqrt{3}}{5} -\\sqrt{3} \\right) -\\left( \\dfrac{1}{5} -\\dfrac{1}{3} \\right) \\\\\r\n& = \\underline{\\dfrac{4 \\sqrt{3}}{5} +\\dfrac{2}{15}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088. (1)\u3000\u4e0d\u5b9a\u7a4d\u5206 \\(\\displaystyle\\int e^{-x} \\sin^2 x \\, dx\\) \u3092\u6c42\u3081\u3088. (2)\u3000\u5b9a\u7a4d\u5206 \\(\\displaystyle\\int _ 0^1 \\sqrt{1 … \u7d9a\u304d\u3092\u8aad\u3080