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Calculating the integral of a function of the form $\\frac{a}{\\sqrt{b^2-x^2}}$ using a table of integrals.

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Using the Table of Integrals/Antiderivatives, calculate the integral of $f(x)=\\simplify[unitFactor]{{a}/(sqrt({b^2}-x^2))}.$

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From the Table of Integrals we see that a function of the form \\[f(x)=\\simplify[unitFactor]{1/(sqrt(a^2-x^2))}\\] has the integral \\[ \\int\\simplify[unitFactor]{1/(sqrt(a^2-x^2))} \\,dx  = \\simplify{arcsin(x/a)+c},\\]

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(Reminder: $\\arcsin(x) \\equiv \\sin^{-1}(x)$)

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and \\[ \\int kf(x) \\,dx = k \\int f(x) \\, dx.\\]

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So, for the function

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\\[f(x)=\\simplify[unitFactor]{{a}/(sqrt({b^2}-x^2))},\\]

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the integral is

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\\[ \\int\\simplify[unitFactor]{{a}/(sqrt({b^2}-x^2))} \\,dx \\,=\\simplify[unitFactor]{{a}int(1/(sqrt({b^2}-x^2)),x)} \\,=\\simplify[all]{{a}(arcsin(x/{b}))} +c.  \\]

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