Why a Line Can Rationalize a Quadratic Root
A square root of a quadratic polynomial can be viewed as a coordinate on a curve. For example, setting \(y=\sqrt{x^2+1}\) places the point \((x,y)\) on the conic \(y^2=x^2+1\). If that conic can be expressed using rational functions of a parameter \(t\), then both \(x\) and \(y\), and also \(dx/dt\), are rational functions of \(t\). This can turn an integral involving the square root into an integral of a rational function.
A useful way to find such a parametrization is to choose a known point on the conic and draw a line through it with variable slope. One intersection is the chosen point; solving for the other intersection gives the parameter formulas.
Deriving the Parametrization
Write a general affine conic as
and suppose the conic contains a known point \((x_0,y_0)\). Introduce coordinates measured from that point:
Since \((x_0,y_0)\) lies on the conic, the constant term in the translated equation is zero. The equation therefore has the form
where \(p=2Ax_0+By_0+D\) and \(q=Bx_0+2Cy_0+E\). A nonvertical line through the chosen point with slope \(t\) is \(v=tu\). Substituting this line gives
The factor \(u=0\) is the known intersection \((x_0,y_0)\). If \(A+Bt+Ct^2\ne 0\), the other root is \(u=-(p+qt)/(A+Bt+Ct^2)\). Then \(v=tu\), yielding the parametrization below.
The factorization also identifies important special cases. If \(p+qt=0\) while \(A+Bt+Ct^2\ne0\), the substituted equation is \((A+Bt+Ct^2)u^2=0\). The line is tangent at the chosen point: the intersection there is repeated, so there is no distinct second point. If both \(p+qt=0\) and \(A+Bt+Ct^2=0\), the substituted equation is identically zero. The entire line then lies on the conic, which can happen when the conic is degenerate. If the denominator is zero but the numerator is not, there is no finite second intersection given by this formula. In each case, the displayed rational formula cannot be used as if it were an ordinary second intersection.
Worked Example: Parametrizing the Unit Circle
Worked Example: The Circle from a Fixed Point
Parametrize \(x^2+y^2=1\) using the known point \((-1,0)\). Let \(u=x+1\) and \(v=y\). Substituting \(x=-1+u\) and \(y=v\) into the circle gives
Use the line \(v=tu\). Then
The root \(u=0\) gives the fixed point. Since \(1+t^2\ne0\) for every real \(t\), the second root is \(u=2/(1+t^2)\). Thus \(v=2t/(1+t^2)\), and
Check the parametrization by substitution:
For every real \(t\), this gives a point on the circle other than \((-1,0)\). Conversely, every circle point other than \((-1,0)\) lies on exactly one nonvertical line through \((-1,0)\), with slope \(t=y/(x+1)\), so the formulas cover all those points. For \(t>0\), \(y>0\), and the parametrization covers the open upper semicircle: it excludes both endpoints. Allowing \(t=0\) includes \((1,0)\); the other endpoint \((-1,0)\) is approached as \(t\) tends to either positive or negative infinity.
From a Conic Parametrization to an Integral
Suppose the integrand involves \(y=\sqrt{q(x)}\), where \(q\) is quadratic. The equation \(y^2=q(x)\) defines a conic. If a suitable point on that conic is available, the line method may express \(x=x(t)\) and \(y=y(t)\) rationally. On an interval where the parametrization is reversible and \(dx/dt\) exists, substitution gives
When \(R\) is rational in its two arguments and the parametrization is rational, the new integrand is rational in \(t\). This observation explains why the method is useful; it does not guarantee that every resulting rational integral is simple. The parameter range and the chosen sign of the square root must also match the interval of the original integral.
Worked Example: Integrating a Hyperbola Radical
Worked Example: Integrating \(1/\sqrt{x^2+1}\)
For \(y^2=x^2+1\), choose the point \((0,1)\) and set \(y=1+tx\). Substitution into the conic gives
The second intersection is \(x=2t/(1-t^2)\), provided \(t^2\ne1\). Its \(y\)-coordinate is
On the upper branch \(y=\sqrt{x^2+1}\), the parameter range \(-1<t<1\) covers all real \(x\). Indeed, \(x=2t/(1-t^2)\) has derivative \(2(1+t^2)/(1-t^2)^2>0\), and tends to negative and positive infinity at the two respective ends of that interval. Differentiating gives
Therefore, for \(-1<t<1\),
Here \(1+t\) and \(1-t\) are positive on the parameter interval. To check the antiderivative in the original variable, use \(t=x/(y+1)\), where \(y=\sqrt{x^2+1}\). Differentiating this expression gives \(dt/dx=1/[y(y+1)]\), and direct simplification gives \(1-t^2=2/(y+1)\). Hence
This agrees with the original integrand.
Worked Example: A Different Quadratic Radical
Worked Example: Integrating \(1/(x\sqrt{x^2+2x+4})\)
Set \(y=\sqrt{x^2+2x+4}\), so \(y^2=x^2+2x+4\), and use the known point \((0,2)\). The line \(y=2+tx\) gives
For \(t^2\ne1\), the second intersection and its \(y\)-coordinate are
On the positive branch \(y=\sqrt{x^2+2x+4}\), \(x(t)\) is defined for \(-1<t<1\), and
For \(x\ne0\), which corresponds to \(t\ne1/2\), substitution now gives
Thus, on any interval that avoids \(x=0\),
To check, differentiate with respect to \(t\): the derivative of the proposed expression is \(1/(2t-1)\), exactly the transformed integrand \(dx/(xy\,dt)\). Since \(dx/dt\ne0\) for \(-1<t<1\), this verifies the derivative with respect to \(x\) as well. The two intervals \(x<0\) and \(x>0\) require separate constants because the original integrand is undefined at \(x=0\).
Common Pitfalls and Practical Checks
A common error is to treat every exceptional slope as a tangent. For instance, in the translated equation \(u[(A+Bt+Ct^2)u+p+qt]=0\), tangency at the chosen point occurs when \(p+qt=0\) and \(A+Bt+Ct^2\ne0\): the equation then has the repeated root \(u=0\). If both expressions vanish, the entire line satisfies the conic equation; it is a component of a degenerate conic, not an ordinary tangent with a distinct second intersection.
Another error is to claim that a parameter interval covers an endpoint merely because the corresponding coordinate has a limit there. In the circle example, \(t>0\) produces \(y>0\) but excludes both endpoints: \(t=0\) gives \((1,0)\), while \((-1,0)\) is not attained at any finite \(t\). State the actual range, not just the limiting behavior.
Before using a parametrization in an integral, check that the point lies on the conic, substitute the proposed formulas back into the conic equation, identify where the denominator vanishes, and determine the parameter interval for the desired branch. After integrating, differentiate the result or compare its differential with the transformed integrand. These checks catch sign, branch, and domain errors.
Check Your Understanding
Answer these questions, keeping track of both algebra and parameter domains.
- In the general conic derivation, what are \(p\) and \(q\), and why does the translated equation have no constant term?
- What conditions on \(p+qt\) and \(A+Bt+Ct^2\) identify tangency at the chosen point? What does it mean if both vanish?
- For the circle parametrization from \((-1,0)\), which point is omitted for finite real \(t\), and what points are covered when \(t>0\)?
- Why does a rational parametrization of \(y^2=q(x)\) turn an integrand rational in \(x\) and \(y\) into a rational function of the parameter?
- Why must the parameter interval be checked before using the substitution in an integral involving a square root?
