Integrating the Square Root of x²−a²

The Secant Substitution for a Radical of a Difference

When the radicand is \(x^2-a^2\), the identity \(\sec^2\theta-1=\tan^2\theta\) suggests \(x=a\sec\theta\). Then \(x^2-a^2=a^2(\sec^2\theta-1)=a^2\tan^2\theta\), and the radical becomes \(a\tan\theta\) for \(0\le\theta<\frac\pi2\). The integral \(\int\sqrt{x^2-a^2}\,dx\) is defined for \(|x|\ge a\), and this tutorial treats the branch \(x\ge a\); the branch \(x\le-a\) follows by symmetry.

Deriving the Formula

Let \(x=a\sec\theta\) with \(0\le\theta<\frac\pi2\). Then \(dx=a\sec\theta\tan\theta\,d\theta\) and \(\sqrt{x^2-a^2}=a\tan\theta\). Hence

The integral of \(\sec^3\theta\) is \(\frac12\bigl(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\bigr)\), and the integral of \(\sec\theta\) is \(\ln|\sec\theta+\tan\theta|\). Subtracting,

From the triangle with hypotenuse \(x\), adjacent side \(a\), and opposite side \(\sqrt{x^2-a^2}\), we read \(\sec\theta=\frac xa\) and \(\tan\theta=\frac{\sqrt{x^2-a^2}}{a}\). Then \(\sec\theta\tan\theta=\frac{x\sqrt{x^2-a^2}}{a^2}\) and \(\sec\theta+\tan\theta=\frac{x+\sqrt{x^2-a^2}}{a}\). The constant \(\ln a\) from the denominator is absorbed in \(C\), which produces the stated result.

Checking by Differentiation

The derivative of \(\frac x2\sqrt{x^2-a^2}\) is \(\frac12\sqrt{x^2-a^2}+\frac{x^2}{2\sqrt{x^2-a^2}}\). The derivative of \(\ln(x+\sqrt{x^2-a^2})\) is \(\frac{1}{\sqrt{x^2-a^2}}\), so the second term contributes \(-\frac{a^2}{2\sqrt{x^2-a^2}}\). Over the common denominator \(2\sqrt{x^2-a^2}\), the numerator is \((x^2-a^2)+x^2-a^2=2(x^2-a^2)\), and the sum is \(\sqrt{x^2-a^2}\).

Worked Examples

A Reliable Procedure

Common Mistakes

Students often forget the minus sign between the two terms, write \(+\frac{a^2}{2}\ln\), or drop the factor \(\frac12\) in front of \(\frac{a^2}{2}\). The sign matters: the sum \(\frac x2\sqrt{x^2-a^2}+\frac{a^2}{2}\ln(\ldots)\) belongs to the plus-sign radical \(\sqrt{x^2+a^2}\), not to \(\sqrt{x^2-a^2}\). A second error is to apply the formula for \(x<a\), where the radicand is negative.

Quick Reference

Interpreting the Result

The curve \(y=\sqrt{x^2-a^2}\) is the upper branch of the hyperbola \(x^2-y^2=a^2\). The area under it from \(x=a\) to \(x\) equals the area of the triangle with vertices \((0,0)\), \((x,0)\), and \((x,y)\), which is \(\frac12xy=\frac x2\sqrt{x^2-a^2}\), minus the area of the hyperbolic sector between the origin and the curve. The sector area is \(\frac{a^2}{2}\ln\frac{x+\sqrt{x^2-a^2}}{a}\). This is the hyperbolic analogue of the sector-plus-triangle picture for the circle, with a subtraction instead of an addition.

A Numerical Spot Check

Test \(\int_1^2\sqrt{x^2-1}\,dx\approx1.074\) with Simpson's rule using three points. The heights are \(0\), \(\sqrt{1.25}\approx1.1180\), and \(\sqrt3\approx1.7321\). Then \(\frac{0.5}{3}(0+4\cdot1.1180+1.7321)\approx1.034\). Simpson's rule is slightly low because the curve has a vertical tangent at \(x=1\), which slows convergence, but the estimate is within four percent.

Practice Problems with Answers

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

1. Evaluate \(\int_1^2\sqrt{x^2-1}\,dx\). Hint: the lower limit gives \(0\). Answer: \(\sqrt3-\frac12\ln(2+\sqrt3)\)
2. Evaluate \(\int_2^4\sqrt{x^2-4}\,dx\). Hint: combine the two logarithms. Answer: \(4\sqrt3-2\ln(2+\sqrt3)\)
3. Find an antiderivative of \(\sqrt{x^2-9}\) for \(x>3\). Hint: use \(a^2=9\). Answer: \(\frac x2\sqrt{x^2-9}-\frac92\ln(x+\sqrt{x^2-9})\)
4. Evaluate \(\int_4^5\sqrt{x^2-16}\,dx\). Hint: at \(x=5\), \(\sqrt{25-16}=3\) and \(x+\sqrt{\cdot}=8\). Answer: \(\frac{15}{2}-8\ln2\)

Recalling the Secant-Cubed Integral

The derivation uses \(\int\sec^3\theta\,d\theta=\frac12\bigl(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\bigr)+C\), which comes from integration by parts. Write \(\sec^3\theta=\sec\theta\cdot\sec^2\theta\), take \(u=\sec\theta\) and \(dv=\sec^2\theta\,d\theta\), and obtain \(\sec\theta\tan\theta-\int\sec\theta\tan^2\theta\,d\theta=\sec\theta\tan\theta-\int\sec^3\theta\,d\theta+\int\sec\theta\,d\theta\). Solving for the integral gives the stated formula. If you forget it, you can rederive it in a minute, and the overall integral then follows.

An Alternative Route with Hyperbolic Functions

Another substitution is \(x=a\cosh t\), with \(t\ge0\). Then \(\sqrt{x^2-a^2}=a\sinh t\) and \(dx=a\sinh t\,dt\), so the integral becomes \(a^2\int\sinh^2t\,dt=\frac{a^2}{2}\bigl(\sinh t\cosh t-t\bigr)\). Since \(\cosh t=\frac xa\) and \(\sinh t=\frac{\sqrt{x^2-a^2}}{a}\), and \(t=\ln\frac{x+\sqrt{x^2-a^2}}{a}\), the result matches the trigonometric answer exactly. The hyperbolic route avoids the \(\sec^3\) integral, but trigonometric substitution remains the standard method in this course.

The Branch x ≤ −a

For \(x\le-a\), the radical \(\sqrt{x^2-a^2}\) is still real. The formula needs an absolute value: \(\frac x2\sqrt{x^2-a^2}-\frac{a^2}{2}\ln|x+\sqrt{x^2-a^2}|\) is valid on both branches. On the left branch, \(x+\sqrt{x^2-a^2}\) is negative, so the absolute value is needed to keep the logarithm defined. In practice, most problems use \(x\ge a\), but a definite integral on a negative interval should be handled with the absolute value or by the substitution \(x\mapsto-x\), which shows that the integral over \([-b,-a]\) equals the integral over \([a,b]\).

Why the Answer Is Not an Inverse Trigonometric Function

For \(\sqrt{a^2-x^2}\) the second term was an arcsine, but here it is a logarithm. The reason is that the secant integral is logarithmic, while the cosine integral produced an angle. The circle is described by angles, and the hyperbola by logarithms. The pairing mirrors the identities \(\cos^2\theta+\sin^2\theta=1\) and \(\cosh^2t-\sinh^2t=1\). Recognizing which curve is behind the radical tells you the type of answer to expect before you start.

Large-x Behavior as a Check

For large \(x\), \(\sqrt{x^2-a^2}\approx x-\frac{a^2}{2x}\), so the integral should behave like \(\frac{x^2}{2}-\frac{a^2}{2}\ln x\). The antiderivative agrees: \(\frac x2\sqrt{x^2-a^2}\approx\frac{x^2}{2}-\frac{a^2}{4}\) and \(\ln(x+\sqrt{x^2-a^2})\approx\ln(2x)\), giving \(\frac{x^2}{2}-\frac{a^2}{2}\ln x\) plus constants. This asymptotic test catches a wrong sign or a wrong coefficient on the logarithm without any differentiation, which is why it is worth running whenever a result looks suspicious.

Where This Fits in the Course

This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Deriving the Secant Substitution for x²−a²,” and the one that follows is “Integrating the Reciprocal Square Root of x²−a².” It is the secant counterpart to the sine substitution of the earlier tutorials, and its logarithmic answer sets the pattern for the reciprocal and polynomial-weighted forms that come next in this module.