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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)}\\]

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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 $\\displaystyle \\simplify[std]{(sqrt({b})/sqrt({a}))*x}$

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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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\\[I=\\simplify[std]{Int(({c} / (sqrt({a}-{b}x^2))),x)}\\]

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$I=\\;$[[0]]

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Input all numbers as integers, fractions or surds. No decimal numbers. You input surds, for example, $\\sqrt{2}$ by writing sqrt(2).

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Input the constant of integration as $C$.

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You can get help by clicking on Show steps. You will lose 1 mark if you do so.

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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 the following integral.

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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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