Integrating the Reciprocal Square Root of x²−a²

The Reciprocal of a Radical of a Difference

The integral \(\int\frac{dx}{\sqrt{x^2-a^2}}\) is the secant-substitution analogue of the arcsine integral. It differs in an important way: the answer is a logarithm rather than an inverse trigonometric function. The integrand is defined for \(|x|>a\), a domain with two disconnected pieces, and the antiderivative has to respect both.

Deriving the Formula

Take the branch \(x>a\). Let \(x=a\sec\theta\) with \(0<\theta<\frac\pi2\). Then \(dx=a\sec\theta\tan\theta\,d\theta\) and \(\sqrt{x^2-a^2}=a\tan\theta\). The integral becomes

From the triangle with hypotenuse \(x\), adjacent side \(a\), and opposite side \(\sqrt{x^2-a^2}\), we have \(\sec\theta=\frac xa\) and \(\tan\theta=\frac{\sqrt{x^2-a^2}}{a}\). So the result is \(\ln\frac{x+\sqrt{x^2-a^2}}{a}+C\). The constant \(-\ln a\) is absorbed into \(C\), giving the stated formula.

Checking by Differentiation

Differentiate \(\ln(x+\sqrt{x^2-a^2})\) for \(x>a\). The chain rule gives \(\frac{1+\frac{x}{\sqrt{x^2-a^2}}}{x+\sqrt{x^2-a^2}}\). Writing the numerator as \(\frac{\sqrt{x^2-a^2}+x}{\sqrt{x^2-a^2}}\), the factor \(x+\sqrt{x^2-a^2}\) cancels with the denominator, leaving \(\frac{1}{\sqrt{x^2-a^2}}\).

Both Branches

For \(x<-a\), the quantity \(x+\sqrt{x^2-a^2}\) is negative, so the absolute value is needed. The derivative of \(\ln|x+\sqrt{x^2-a^2}|\) is the same expression as before, \(\frac{1}{\sqrt{x^2-a^2}}\), because \(\frac{d}{dx}\ln|g|=\frac{g'}{g}\) regardless of the sign of \(g\). Thus the formula holds on both branches, though the constant \(C\) may differ from one branch to the other. A definite integral must lie entirely within one branch.

Worked Examples

A Reliable Procedure

Common Mistakes

A frequent mistake is to write \(\ln|x-\sqrt{x^2-a^2}|\) or to omit the radical altogether, producing \(\ln|x|\). Another is to confuse the result with the arcsecant: \(\int\frac{dx}{x\sqrt{x^2-a^2}}\) has an arcsecant, but \(\int\frac{dx}{\sqrt{x^2-a^2}}\) does not. A third mistake is to ignore the domain: the formula does not apply for \(|x|<a\), where the radicand is negative. Finally, do not drop the scale factor: \(\int\frac{dx}{\sqrt{4x^2-1}}\) is half of the logarithm, and leaving out the \(\frac12\) doubles the answer. A quick differentiation exposes the error, because the chain rule produces an extra factor of \(2\) from the inner function \(2x\), which the outside \(\frac12\) must cancel.

Quick Reference

Interpreting the Result

The antiderivative \(\ln(x+\sqrt{x^2-a^2})\) equals \(\operatorname{arccosh}\frac xa+\ln a\), the inverse hyperbolic cosine. This matches the geometric meaning: as the circle was parametrized by angle, the hyperbola \(x^2-y^2=a^2\) is parametrized by hyperbolic angle, and the integral of the arc-length-like element \(\frac{dx}{\sqrt{x^2-a^2}}\) measures that parameter. For large \(x\), \(\ln(x+\sqrt{x^2-a^2})\approx\ln(2x)\), so the integral grows logarithmically, much more slowly than \(x\).

A Numerical Spot Check

Check the antiderivative on the sub-interval \([4,5]\), away from the singularity at \(x=3\). The exact value is \(\ln9-\ln(4+\sqrt7)\approx0.3032\). The midpoint rule with one rectangle uses the height at \(x=4.5\), namely \(\frac{1}{\sqrt{11.25}}\approx0.2981\), and gives \(0.298\). The estimate is slightly low because the integrand is convex and decreasing, and it agrees with the exact value to within two percent. The full integral \(\ln3\approx1.099\) is larger because the part over \([3,4]\) contributes about \(0.796\), where the integrand blows up near \(x=3\).

Practice Problems with Answers

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

1. Evaluate \(\int_1^2\frac{dx}{\sqrt{x^2-1}}\). Hint: the lower limit gives \(\ln1=0\). Answer: \(\ln(2+\sqrt3)\)
2. Evaluate \(\int_3^5\frac{dx}{\sqrt{x^2-9}}\). Hint: use the 3-4-5 triangle. Answer: \(\ln3\)
3. Find an antiderivative of \(\frac{1}{\sqrt{4x^2-1}}\) for \(x>\frac12\). Hint: let \(u=2x\). Answer: \(\frac12\ln(2x+\sqrt{4x^2-1})\)
4. Evaluate \(\int_{-5}^{-3}\frac{dx}{\sqrt{x^2-9}}\). Hint: use the absolute value or symmetry. Answer: \(\ln3\)

The Hyperbolic Form

With \(x=a\cosh t\), \(t>0\), we have \(\sqrt{x^2-a^2}=a\sinh t\) and \(dx=a\sinh t\,dt\), so the integrand reduces to \(\int dt=t=\operatorname{arccosh}\frac xa\). This is the same function as \(\ln\bigl(\frac{x+\sqrt{x^2-a^2}}{a}\bigr)\). Textbooks often write the answer either way; the logarithmic form needs no special functions and is the one expected here.

Contrast with the Arcsine Integral

The comparison shows that the two integrals live on complementary regions of the real line. Inside the circle the integral is bounded by \(\frac\pi2\), but outside it the integral grows without limit, though slowly, like \(\ln x\).

Why the Integral Converges at x = a

Near \(x=a\) the radicand factors as \(x^2-a^2=(x-a)(x+a)\approx2a(x-a)\), so the integrand behaves like \(\frac{1}{\sqrt{2a}\sqrt{x-a}}\). A function of the form \((x-a)^{-1/2}\) is integrable at \(a\), because its antiderivative \(2\sqrt{x-a}\) tends to zero. The logarithmic antiderivative reflects this: \(\ln(x+\sqrt{x^2-a^2})\) is continuous at \(x=a\), with value \(\ln a\). Integrals starting exactly at \(x=a\) are therefore improper but finite, and the usual Fundamental Theorem evaluation \(F(b)-F(a)\) is valid when interpreted as a limit.

Growth for Large x

For large \(x\), \(\sqrt{x^2-a^2}\approx x\), so the integrand is close to \(\frac1x\) and the integral resembles \(\ln x\). More precisely, \(\ln(x+\sqrt{x^2-a^2})\approx\ln(2x)\), which equals \(\ln x+\ln2\). This agrees with the idea that the integral of \(\frac1x\) is \(\ln x\), with a small correction from the radical. The estimate gives a quick consistency check: any proposed antiderivative should grow like a logarithm, not like a power of \(x\).

Choosing the Right Form of the Answer

Some instructors prefer \(\operatorname{arccosh}\frac xa\) and others prefer the logarithm. Because the two differ by the constant \(\ln a\), either is a correct antiderivative, and a definite integral gives the same number from both. On an exam, the logarithmic form is safer, since it uses only functions introduced earlier in the course and is easy to differentiate for verification.

Where This Fits in the Course

This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Integrating the Square Root of x²−a²,” and the one that follows is “Polynomial Factors Over √(x²−a²).” It pairs with the arcsine integral from earlier: the sine substitution gives an angle, and the secant substitution gives a logarithm. The next tutorials combine this reciprocal form with polynomial and reciprocal-variable factors, so the logarithmic antiderivative and the domain conditions stated here will reappear in each of them and should be second nature before you continue.