\u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
(1)<\/strong>\u3000\u4e0d\u5b9a\u7a4d\u5206\r\n\\[\r\n\\displaystyle\\int \\sqrt{1 -e^{-2x}} \\, dx\r\n\\]\r\n\u3092\u7f6e\u63db \\(\\sqrt{1 -e^{-2x}} = t\\) \u3092\u7528\u3044\u3066\u6c42\u3081\u3088.<\/p><\/li>\r\n (2)<\/strong>\u3000\u6975\u9650\r\n\\[\r\n\\displaystyle\\lim _ {\\alpha \\rightarrow \\infty} \\displaystyle\\int _ 0^{\\alpha} \\left( 1 -\\sqrt{1 -e^{-2x}} \\right) \\, dx\r\n\\]\r\n\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(\\sqrt{1 -e^{-2x}} = t\\) \u3088\u308a\r\n\\[\\begin{align}\r\ne^{-2x} & = 1-t^2 \\\\\r\n\\text{\u2234} \\quad x & = -\\dfrac{1}{2} \\log ( 1-t^2 )\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n\\dfrac{dx}{dt} = -\\dfrac{-2t}{2 (1-t^2)} = \\dfrac{t}{1-t^2}\r\n\\]\r\n\u3088\u3063\u3066\r\n\\[\\begin{align}\r\n\\displaystyle\\int \\sqrt{1 -e^{-2x}} \\, dx & =\\displaystyle\\int t \\cdot \\dfrac{t}{1-t^2} \\, dt \\\\\r\n& = \\displaystyle\\int \\left( \\dfrac{1}{1-t^2} -1 \\right) \\, dt \\\\\r\n& = \\displaystyle\\int \\left\\{ \\dfrac{1}{2(1+t)} +\\dfrac{1}{2(1-t)} -1 \\right\\} \\, dt \\\\\r\n& = \\dfrac{1}{2} \\log |1+t| -\\dfrac{1}{2} \\log |1-t| -t +C \\\\\r\n& = \\dfrac{1}{2} \\log \\left| \\dfrac{1 +\\sqrt{1 -e^{-2x}}}{1 -\\sqrt{1 -e^{-2x}}} \\right| -\\sqrt{1 -e^{-2x}} +C \\\\\r\n& = \\dfrac{1}{2} \\log \\left| \\dfrac{\\left( 1 +\\sqrt{1 -e^{-2x}} \\right)^2}{e^{-2x}} \\right| -\\sqrt{1 -e^{-2x}} +C \\\\\r\n& = \\underline{\\log \\left( 1 +\\sqrt{1 -e^{-2x}} \\right) +x -\\sqrt{1 -e^{-2x}} +C}\r\n\\end{align}\\]\r\n\u305f\u3060\u3057, \\(C\\) \u306f\u7a4d\u5206\u5b9a\u6570.<\/p>\r\n (2)<\/strong><\/p>\r\n (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\\begin{align}\r\n\\displaystyle\\int _ 0^{\\alpha} & \\left( 1 -\\sqrt{1 -e^{-2x}} \\right) \\, dx \\\\\r\n& = \\left[ \\sqrt{1 -e^{-2x}} -\\log \\left( 1 +\\sqrt{1 -e^{-2x}} \\right) \\right] _ 0^{\\alpha} \\\\\r\n& = \\sqrt{1 -e^{-2\\alpha}} -\\log \\left( 1 +\\sqrt{1 -e^{-2\\alpha}} \\right) \\\\\r\n& \\rightarrow \\sqrt{1} -\\log \\left( 1 +\\sqrt{1} \\right) \\quad ( \\alpha \\rightarrow \\infty \\text{\u306e\u3068\u304d} ) \\\\\r\n& = \\underline{1-\\log 2}\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 \\sqrt{1 -e^{-2x}} \\, dx \\] \u3092\u7f6e\u63db \\(\\sqrt{1 -e^{-2x}} = t\\) \u3092\u7528\u3044\u3066\u6c42\u3081\u3088. (2)\u3000 … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n