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For the integral \\[I=\\simplify[std]{Int((({c}) / (sqrt({a}-{b}x^2))),x)}\\] use the substitution $\\displaystyle \\simplify[std]{u=(sqrt({b})/sqrt({a}))*x}$
so that \\[\\simplify[all,!sqrtProduct,fractionNumbers]{sqrt({a}-{b}x^2)=sqrt({a}-{b}*({a}/{b})*u^2)=sqrt({a}-{a}*u^2)=sqrt({a})*sqrt(1-u^2)}\\]
We have $\\displaystyle \\simplify[std]{du=(sqrt({b})/sqrt({a}))dx}$ and we get
\\[\\begin{eqnarray*}I&=&\\simplify[std]{({c}*(sqrt({a})/sqrt({b})))*Int((1 / ( sqrt({a})*sqrt(1-u^2) )),u)}\\\\ &=&\\simplify[std]{({c}/sqrt({b}))*Int((1 / (sqrt(1-u^2))),u)}\\\\ &=&\\simplify[std]{({c}/sqrt({b}))*arcsin(u)+C}\\\\ &=&\\simplify[std]{({c}/sqrt({b}))*arcsin((sqrt({b})/sqrt({a}))*x)+C} \\end{eqnarray*}\\]
on replacing $u$ by $\\displaystyle \\simplify[std]{(sqrt({b})/sqrt({a}))*x}$
Try the substitution $\\displaystyle \\simplify[std]{u=(sqrt({b})/sqrt({a}))*x}$ and then consider the standard integral \\[\\int \\frac{dx}{\\sqrt{1-x^2}}=\\arcsin(x)+C\\]
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\n\t\t\t$I=\\;$[[0]]
\n\t\t\tInput all numbers as integers, fractions or surds. No decimal numbers. You input surds, for example, $\\sqrt{2}$ by writing sqrt(2).
\n\t\t\tInput the constant of integration as $C$.
\n\t\t\tYou can get help by clicking on Show steps. You will lose 1 mark if you do so.
\n\t\t\t", "showFeedbackIcon": true, "gaps": [{"checkVariableNames": false, "customMarkingAlgorithm": "", "showPreview": true, "checkingType": "absdiff", "type": "jme", "vsetRangePoints": 5, "scripts": {}, "answerSimplification": "std", "unitTests": [], "showCorrectAnswer": true, "variableReplacements": [], "failureRate": 1, "answer": "({c}/sqrt({b}))*arcsin((sqrt({b})/sqrt({a}))*x)+C", "marks": 3, "checkingAccuracy": 0.001, "notallowed": {"partialCredit": 0, "showStrings": false, "strings": ["."], "message": "Do not input numbers as decimals, only as integers without the decimal point, or fractions or surds (such as sqrt(2) for $\\sqrt{2}$).
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Find $\\displaystyle \\int \\frac{c}{\\sqrt{a-bx^2}}\\;dx$. Solution involves the inverse trigonometric function $\\arcsin$.
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