Polynomial Factors Over √(x²−a²)

Polynomial Weights Over a Radical

The previous tutorial integrated \(\frac{1}{\sqrt{x^2-a^2}}\). Here the numerator carries a power of \(x\): the integrals \(\int\frac{x^n}{\sqrt{x^2-a^2}}\,dx\) for \(n=1,2,3\). The secant substitution \(x=a\sec\theta\) handles all of them, and it turns each integral into a power of secant. For odd \(n\) there is an additional shortcut, the substitution \(u=\sqrt{x^2-a^2}\), which avoids trigonometry entirely.

Deriving the Three Formulas

Let \(x=a\sec\theta\) with \(0\le\theta<\frac\pi2\), so that \(dx=a\sec\theta\tan\theta\,d\theta\) and \(\sqrt{x^2-a^2}=a\tan\theta\). Then \(\frac{dx}{\sqrt{x^2-a^2}}=\sec\theta\,d\theta\), and \(\frac{x^n\,dx}{\sqrt{x^2-a^2}}=a^n\sec^{n+1}\theta\,d\theta\). The three cases are

The first gives \(a\tan\theta=\sqrt{x^2-a^2}\). The second uses \(\int\sec^3\theta\,d\theta=\frac12(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)\), which becomes \(\frac x2\sqrt{x^2-a^2}+\frac{a^2}{2}\ln(x+\sqrt{x^2-a^2})\) after substituting \(\sec\theta=\frac xa\) and \(\tan\theta=\frac{\sqrt{x^2-a^2}}{a}\). The third uses \(\sec^4\theta=\sec^2\theta(1+\tan^2\theta)\), so \(\int\sec^4\theta\,d\theta=\tan\theta+\frac{\tan^3\theta}{3}\), which gives \(a^3\bigl(\frac{\sqrt{x^2-a^2}}{a}+\frac{(x^2-a^2)^{3/2}}{3a^3}\bigr)\), the stated result.

The Algebraic Shortcut for Odd Powers

For odd \(n\), the substitution \(u=\sqrt{x^2-a^2}\) works without trigonometry. Then \(u^2=x^2-a^2\), \(u\,du=x\,dx\), and \(x^2=u^2+a^2\). For \(n=3\), write \(x^3\,dx=x^2\cdot x\,dx=(u^2+a^2)u\,du\), so

This agrees with the trigonometric answer. The shortcut is not available for even \(n\), because the leftover factor \(x\) has no partner to form \(x\,dx\).

A Reduction Relation as a Check

Differentiating \(x^{n-1}\sqrt{x^2-a^2}\) gives \(nx^{n}/\sqrt{x^2-a^2}-(n-1)a^2x^{n-2}/\sqrt{x^2-a^2}\), so

For \(n=2\) this reproduces the second formula, using the logarithmic integral from the previous tutorial. For \(n=3\) it gives \(\frac{x^2\sqrt{x^2-a^2}}{3}+\frac{2a^2}{3}\sqrt{x^2-a^2}\), which equals \(\frac{(x^2+2a^2)\sqrt{x^2-a^2}}{3}\). The same function results from the other form, since \(\frac{(x^2-a^2)}{3}+a^2=\frac{x^2+2a^2}{3}\).

Worked Examples

A Reliable Procedure

Common Mistakes

The first mistake is to forget the factor \(x^n\) in the numerator when converting to \(\theta\), losing the powers of \(a\sec\theta\). The second is to drop the coefficient \(a^2\) in front of the term \(\sqrt{x^2-a^2}\) in the cubic case. A third is to apply the algebraic shortcut \(u=\sqrt{x^2-a^2}\) to even powers, where it leaves behind an extra factor of \(x\).

Practice Problems with Answers

Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.

1. Evaluate \(\int_1^2\frac{x\,dx}{\sqrt{x^2-1}}\). Hint: let \(u=\sqrt{x^2-1}\). Answer: \(\sqrt3\)
2. Evaluate \(\int_2^4\frac{x^2\,dx}{\sqrt{x^2-4}}\). Hint: use the secant-cubed formula. Answer: \(4\sqrt3+2\ln(2+\sqrt3)\)
3. Evaluate \(\int_1^2\frac{x^3\,dx}{\sqrt{x^2-1}}\). Hint: use \(\frac{u^3}{3}+u\) with \(u=\sqrt3\). Answer: \(2\sqrt3\)
4. Evaluate \(\int_3^5\frac{x^3\,dx}{\sqrt{x^2-9}}\). Hint: here \(u=4\) and \(\frac{u^3}{3}+9u\). Answer: \(\frac{172}{3}\)

Interpreting the Results

For large \(x\) the integrand \(\frac{x^n}{\sqrt{x^2-a^2}}\) behaves like \(x^{n-1}\), so the antiderivative should grow like \(\frac{x^n}{n}\). The three formulas agree: the first is about \(x\), the second about \(\frac{x^2}{2}\), and the third about \(\frac{x^3}{3}\). The extra terms, such as the logarithm and the term \(a^2\sqrt{x^2-a^2}\), are lower-order corrections that become relatively small for large \(x\).

Near \(x=a\), the factor \(x^n\) is close to the constant \(a^n\), so the integrand behaves like \(\frac{a^n}{\sqrt{2a}\sqrt{x-a}}\). This is integrable, which explains why all definite integrals starting at \(x=a\) converge, whatever the power \(n\).

A Numerical Spot Check

Check \(\int_1^2\frac{x\,dx}{\sqrt{x^2-1}}=\sqrt3\) with the substitution \(w=x^2-1\), which gives \(\frac12\int_0^3w^{-1/2}\,dw=\frac12\cdot2\sqrt3=\sqrt3\). For a numerical check that avoids the singularity, evaluate the integral over \([1.5,2]\): the exact value is \(\sqrt3-\sqrt{1.25}\approx0.614\), and the midpoint rule with the height at \(x=1.75\), \(\frac{1.75}{\sqrt{2.0625}}\approx1.2187\), gives \(0.609\). The two agree to within one percent.

Choosing the Method

Use the algebraic substitution \(u=\sqrt{x^2-a^2}\) whenever the power of \(x\) is odd, because it is shorter and avoids trigonometric identities. Use the secant substitution for even powers, where the answer necessarily includes a logarithm. If you are unsure, the reduction relation always works and provides a built-in check, because each application can be verified by differentiating one term.

Why Parity of n Matters

The integrand \(\frac{x^n}{\sqrt{x^2-a^2}}\) has a clear parity: it is even when \(n\) is even and odd when \(n\) is odd. In the odd case, the factor \(x\,dx\) is exactly half the differential of \(x^2-a^2\), so the integrand is a function of \(x^2\) times \(x\,dx\). That structure signals a substitution in \(w=x^2-a^2\). In the even case, there is no such pairing, and the antiderivative necessarily contains the logarithm from the \(\sec^3\) integral. The parity therefore predicts the type of the answer before any computation: algebraic for odd \(n\), algebraic plus logarithmic for even \(n\).

Domain Notes

All formulas above are stated for \(x>a\). On the left branch \(x<-a\), the radical is still real, but the substitution \(x=a\sec\theta\) requires \(\theta\) in the second quadrant, where \(\tan\theta<0\). The simplest way to handle the left branch is to substitute \(x=-t\) with \(t>a\), which turns the integral into a multiple of the corresponding right-branch integral, with the sign \((-1)^n\): over an interval \([-b,-a]\), the integral equals \((-1)^n\) times the integral of the same expression over \([a,b]\). Thus the area on the left branch mirrors the right branch for even \(n\) and is the negative of it for odd \(n\). Because most applications use the right branch, we keep the discussion there, but this symmetry lets you reuse every result in the tutorial.

Where This Fits in the Course

This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Integrating the Reciprocal Square Root of x²−a²,” and the one that follows is “Integrals With x√(x²−a²) in the Denominator.” It extends the reciprocal-radical integral by polynomial weights, and the reduction relation introduced here is a preview of the reduction formulas for powers of trigonometric functions that appear later in the course, where each step lowers the power until a basic integral remains.