Why This Parameter Helps
The expression \(\sqrt{a^2-x^2}\) is not rational in \(x\), even though it is built from a quadratic polynomial. A suitable change of variable can make both the radical and the differential rational functions of a new variable. This is useful when the resulting rational integral is easier to evaluate than the original one.
Throughout, assume \(a>0\) and \(-a\le x\le a\), so that \(a^2-x^2\ge 0\) and the square root means the principal, nonnegative square root. The parameter we use is
To handle the principal square root without an absolute value, we restrict the parameter to \(t\ge 0\). On this branch, \(t=0\) corresponds to \(x=a\), and increasing \(t\) moves \(x\) down toward \(-a\). Keeping track of this direction is essential when changing bounds.
Deriving the Rational Formulas
The parameter formula can be obtained directly from the inverse relation. Starting from \(t^2=(a-x)/(a+x)\), multiply by \(a+x\), then collect the terms involving \(x\):
To derive the radical, factor the difference of squares and substitute the parameter formula. Because \(t\ge0\), no absolute-value ambiguity remains:
Differentiating \(x=a(1-t^2)/(1+t^2)\) gives the differential:
These formulas explain why the substitution works. If an integrand is a rational expression in \(x\) and \(\sqrt{a^2-x^2}\), replacing both by the displayed rational functions of \(t\), and multiplying by \(dx\), produces a rational expression in \(t\), wherever the transformed denominator is nonzero.
Worked Integrals
Worked Example: Integrating \(x/\sqrt{a^2-x^2}\)
Assume \(a>0\) and \(-a<x<a\). Apply the parameter to find an antiderivative of \(\frac{x}{\sqrt{a^2-x^2}}\).
First express the quotient and differential in terms of \(t\):
Thus the transformed integrand is
Since \(\frac{d}{dt}\bigl(\frac{t}{1+t^2}\bigr)=\frac{1-t^2}{(1+t^2)^2}\), integration gives
To check, \(\frac{2at}{1+t^2}=\sqrt{a^2-x^2}\), so this is \(-\sqrt{a^2-x^2}+C\). Differentiating with respect to \(x\) gives \(\frac{x}{\sqrt{a^2-x^2}}\), as required.
Worked Example: A Definite Integral with Reversed Parameter Bounds
Evaluate \(\int_0^3 \frac{x}{\sqrt{25-x^2}}\,dx\). Here \(a=5\), and the original interval \([0,3]\) lies inside \((-5,5)\).
At the original endpoints, the nonnegative parameter has values
Using the transformed integrand from the preceding calculation, with these bounds in their correct order, gives
At \(t=1/2\), the antiderivative is \(-10(1/2)/(1+1/4)=-4\). At \(t=1\), it is \(-10/2=-5\). Therefore the integral is \((-4)-(-5)=1\). As a check, the original antiderivative is \(-\sqrt{25-x^2}\); its endpoint difference is \(-4-(-5)=1\).
Worked Example: Integrating the Reciprocal of the Radical
Evaluate \(\int_0^3 \frac{dx}{\sqrt{25-x^2}}\), again using \(a=5\). The parameter bounds are \(t=1\) at \(x=0\) and \(t=1/2\) at \(x=3\).
Substitute the radical and differential:
Therefore,
For a derivative check, \(dt/dx=-(1+t^2)^2/(20t)\), obtained by inverting \(dx/dt=-20t/(1+t^2)^2\). Hence
The antiderivative differentiates to the original integrand, and evaluating it at \(x=0\) and \(x=3\) gives the same endpoint difference shown above.
Worked Example: A Rationalized Integral with a Linear Factor
Evaluate \(\int_0^3 \frac{dx}{(5+x)\sqrt{25-x^2}}\). The parameter bounds are again \(1\) and \(1/2\).
The linear factor and radical become
Including the differential, the transformed integrand is especially simple:
Thus
To check the antiderivative, \(t=\sqrt{(5-x)/(5+x)}\) has derivative \(dt/dx=-5/[t(5+x)^2]\). Therefore \(\frac{d}{dx}(-t/5)=1/[t(5+x)^2]\). Since \(t^2=(5-x)/(5+x)\), this equals \(1/[(5+x)\sqrt{25-x^2}]\). The derivative and the definite value are confirmed.
Bounds, Endpoints, and Orientation
The parameter decreases as \(x\) increases: \(dx/dt<0\) for \(t>0\). Consequently, if the original bounds increase from \(x_1\) to \(x_2\), the corresponding parameter bounds decrease from \(t_1\) to \(t_2\). Keep this reversed order, together with the negative sign in \(dx\), or explicitly reverse the bounds and change the sign—but do not do both.
At \(x=a\), the parameter is \(t=0\), and the formulas remain meaningful by their endpoint limits, though some transformed integrands may need to be interpreted as limits. At \(x=-a\), the inverse formula sends \(t\) to \(+\infty\). An integral with that endpoint must therefore be transformed using a limit as \(t\to+\infty\); it is not correct to substitute a finite parameter value for \(x=-a\).
Confirm \(a>0\) and that the interval lies within \([-a,a]\), where the radical is real.
Use \(t=\sqrt{(a-x)/(a+x)}\), so the radical is \(2at/(1+t^2)\).
Replace \(x\), the radical, and \(dx\), retaining the negative sign in \(dx\).
Compute \(t\) at each original endpoint; handle \(x=-a\) by a limit at infinity.
Differentiate an antiderivative or check the evaluated endpoint difference against the transformed integral.
A Common Branch and Sign Mistake
A tempting but incorrect step is to let \(t\) take either sign and write \(\sqrt{a^2-x^2}=2at/(1+t^2)\). For example, let \(a=5\) and \(t=-1\). The parameter formula gives \(x=0\), so the principal radical is \(\sqrt{25}=5\). But the proposed expression gives \(2(5)(-1)/(1+1)=-5\), which is not the principal square root.
The correction is to restrict to \(t\ge0\). If a signed parameter is used for some separate reason, the correct expression is \(2a|t|/(1+t^2)\), not a rational function across both signs. A second frequent error is changing increasing \(x\)-bounds to increasing \(t\)-bounds without accounting for the fact that \(t\) decreases with \(x\). Compute each endpoint from the inverse formula before integrating.
Why the Substitution Is Useful
This parameter describes the full interior of the interval \((-a,a)\) with a single nonnegative variable and turns the radical into a rational expression. It can therefore convert many algebraic integrals into rational-function integrals, a class of integrals treated by standard algebraic methods. Its usefulness depends on respecting its domain: the branch choice determines the sign of the radical, and the inverse map determines the correct bounds.
The main formulas can be kept together for reference:
They apply for \(a>0\) and \(-a<x\le a\), with the endpoint \(x=-a\) represented by \(t\to+\infty\). For a definite integral, both the branch and the transformed bounds are part of the method, not optional bookkeeping.
Check Your Understanding
Use the branch and interval conditions as well as the algebra when answering these questions.
- For \(a>0\), what value of \(t\ge0\) corresponds to \(x=0\), and what values correspond to \(x=a\) and \(x=-a\)?
- Derive the formula for \(dx\) by differentiating \(x=a(1-t^2)/(1+t^2)\).
- For \(a=4\) and \(x=2\), find \(t\) and verify the rational formula for the principal square root.
- Why is \(\sqrt{a^2-x^2}=2at/(1+t^2)\) not valid for a negative \(t\) without an absolute value?
- When \(x\) increases inside \((-a,a)\), does \(t\) increase or decrease, and what does that imply for definite-integral bounds?
