First, specify the answer to $\\displaystyle{\\int \\frac{1}{\\sqrt{x^2-1}}dx}$:
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\n$I=\\;\\;$[[0]]
\nInput your answer to $2$ decimal places.
", "showCorrectAnswer": true, "marks": 0}], "statement": "In this question the aim is to use hyperbolic functions to find the value of:
\\[I=\\int_{\\var{b}}^{\\var{2*b}} \\left(\\frac{1}{\\sqrt{\\simplify[std]{{a^2}x^2-{b^2}}}}\\right)\\;dx\\]
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\n22/12/2012:(WHF)
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", "licence": "Creative Commons Attribution 4.0 International", "description": "
Find (hyperbolic substitution):
$\\displaystyle \\int_{b}^{2b} \\left(\\frac{1}{\\sqrt{a^2x^2-b^2}}\\right)\\;dx$
\\[I=\\int_{\\var{b}}^{\\var{2*b}} \\left(\\frac{1}{\\sqrt{\\simplify[std]{{a^2}x^2-{b^2}}}}\\right)\\;dx\\]
\nTo solve this, we will employ the standard integral:\\[\\int \\frac{1}{\\sqrt{x^2-1}}\\;dx=\\simplify{arccosh(x)+C}\\]
\nWe wish to rewrite the integrand of $I$ such that it is of the form $\\displaystyle{\\frac{A}{\\sqrt{x^2-1}}}$ ($A$ is a constant) and hence we can use the standard integral. Let us rearrange the integrand into this form:
\n\\[\\begin{eqnarray*}\\frac{1}{\\sqrt{\\simplify[std]{{a^2}x^2-{b^2}}}} &=& \\frac{1}{\\sqrt{\\simplify[std]{({a}x)^2-{b^2}}}} = \\frac{1}{\\sqrt{\\simplify[std]{{b^2}(({a}x/{b})^2-1)}}} \\\\ &=& \\frac{1}{\\var{b}\\sqrt{\\simplify[std]{(({a}x/{b})^2-1)}}} = \\frac{1}{\\var{b}\\sqrt{\\simplify[std]{(v^2-1)}}}\\end{eqnarray*} \\]
\nwhere we have made the substitution $\\displaystyle{v = \\simplify[std]{{a}x/{b}}}$. Following this then $\\displaystyle{dx = \\simplify[std]{{b}/{a}}}dv$. We can then evaluate our integral in terms of $v$ (being careful to modify their limits of integration from $x$ to $v$):
\n\\[\\begin{eqnarray*} I &=& \\int_{\\var{a}}^{\\var{2*a}}\\frac{1}{\\var{b}\\sqrt{v^2-1}}\\simplify[std]{{b}/{a}}dv \\\\&=&\\frac{1}{\\var{a}}\\int_{\\var{a}}^{\\var{2*a}}\\frac{1}{\\sqrt{v^2-1}}\\;dv\\\\ &=&\\frac{1}{\\var{a}}\\left[\\simplify{arccosh(v)}\\right]_{\\var{a}}^{\\var{2*a}}\\\\ &=&\\var{val} \\end{eqnarray*} \\] to 2 decimal places.
", "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}]}], "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}]}