The Arcsecant Integral
The integral \(\int\frac{dx}{x\sqrt{x^2-a^2}}\) combines the radical \(\sqrt{x^2-a^2}\) with a reciprocal factor of \(x\). It is the integral that produces the inverse secant, much as \(\int\frac{dx}{\sqrt{a^2-x^2}}\) produced the inverse sine. The substitution \(x=a\sec\theta\) turns the radical into \(a\tan\theta\) and the differential into \(a\sec\theta\tan\theta\,d\theta\), and then almost everything cancels.
Deriving the Formula
Take the branch \(x>a\) and let \(x=a\sec\theta\) with \(0<\theta<\frac\pi2\). Then \(dx=a\sec\theta\tan\theta\,d\theta\) and \(\sqrt{x^2-a^2}=a\tan\theta\), so
The secant and tangent factors cancel completely, leaving only the angle. The angle is read directly from the substitution: \(\theta=\operatorname{arcsec}\frac xa\), the angle in the triangle with hypotenuse \(x\) and adjacent side \(a\).
The Branch x < −a
On the left branch, take \(\theta\) in \((\frac\pi2,\pi)\), where \(\sec\theta<0\) and \(\tan\theta<0\). Then \(\sqrt{x^2-a^2}=a|\tan\theta|=-a\tan\theta\), and the integrand reduces to \(-\frac1a\,d\theta\). The antiderivative is \(-\frac\theta a\), which equals \(\frac1a\operatorname{arcsec}\frac{|x|}{a}\) up to a constant, because \(\theta=\pi-\operatorname{arcsec}\frac{|x|}{a}\) there. The single formula with \(|x|\) therefore covers both branches. The sign of the integrand tracks the sign of \(x\), and the function \(\operatorname{arcsec}\frac{|x|}{a}\) is even, so its derivative is odd, matching the odd integrand.
An Algebraic Cross-Check
For \(x>a\), set \(t=\sqrt{x^2-a^2}\). Then \(t\,dt=x\,dx\) and \(x^2=t^2+a^2\), so
Hence the antiderivative can also be written as \(\frac1a\arctan\frac{\sqrt{x^2-a^2}}{a}\). In the triangle with hypotenuse \(x\) and adjacent side \(a\), the opposite side is \(\sqrt{x^2-a^2}\), so \(\tan\theta=\frac{\sqrt{x^2-a^2}}{a}\), and the two forms are the same angle. The tangent form is convenient when your calculator has no arcsecant key.
Checking by Differentiation
The derivative of \(\operatorname{arcsec}u\) is \(\frac{1}{|u|\sqrt{u^2-1}}\). With \(u=\frac xa\) and the chain-rule factor \(\frac1a\), the derivative of \(\frac1a\operatorname{arcsec}\frac xa\) for \(x>a\) is \(\frac1a\cdot\frac{1}{\frac xa\sqrt{\frac{x^2}{a^2}-1}}\cdot\frac1a=\frac{1}{x\sqrt{x^2-a^2}}\), as required.
Worked Examples
Worked Example: Radius One
Evaluate \(\int_{\sqrt2}^{2}\frac{dx}{x\sqrt{x^2-1}}\).
The antiderivative is \(\operatorname{arcsec}x\). At \(x=2\): \(\frac\pi3\). At \(x=\sqrt2\): \(\frac\pi4\). Therefore
The integrand is about \(\frac{1}{\sqrt2}=0.707\) at the left endpoint and \(\frac{1}{2\sqrt3}\approx0.289\) at the right, so an area near \(0.26\) over an interval of width \(0.586\) fits.
Worked Example: Radius Two
Evaluate \(\int_{2\sqrt2}^{4}\frac{dx}{x\sqrt{x^2-4}}\).
With \(a=2\), the antiderivative is \(\frac12\operatorname{arcsec}\frac x2\). At \(x=4\): \(\frac12\cdot\frac\pi3\). At \(x=2\sqrt2\): \(\frac12\cdot\frac\pi4\). So
The factor \(\frac1a=\frac12\) multiplies the difference of angles.
Worked Example: An Indefinite Integral with Radius Three
Find \(\int\frac{dx}{x\sqrt{x^2-9}}\) for \(x>3\) and evaluate it from \(3\sqrt2\) to \(6\).
The antiderivative is \(\frac13\operatorname{arcsec}\frac x3\). At \(x=6\): \(\operatorname{arcsec}2=\frac\pi3\). At \(x=3\sqrt2\): \(\operatorname{arcsec}\sqrt2=\frac\pi4\). Hence
Alternatively, the tangent form \(\frac13\arctan\frac{\sqrt{x^2-9}}{3}\) gives \(\frac13[\arctan\sqrt3-\arctan1]=\frac13\cdot\frac\pi{12}\), the same value.
Worked Example: The Left Branch
Evaluate \(\int_{-2}^{-\sqrt2}\frac{dx}{x\sqrt{x^2-1}}\).
The antiderivative is \(\operatorname{arcsec}|x|\). At \(x=-\sqrt2\): \(\frac\pi4\). At \(x=-2\): \(\frac\pi3\). Subtracting,
The answer is negative because the integrand is negative for \(x<0\). This is the mirror image of the first example, with the sign reversed by the odd symmetry.
A Reliable Procedure
Write the radicand as \(x^2-a^2\) with the reciprocal factor \(\frac1x\).
Set \(x=a\sec\theta\); the integrand becomes \(\frac1a\,d\theta\).
Use \(\theta=\operatorname{arcsec}\frac{|x|}{a}\), or equivalently \(\arctan\frac{\sqrt{x^2-a^2}}{a}\).
Multiply by \(\frac1a\).
Differentiate, or compare with the substitution \(t=\sqrt{x^2-a^2}\).
Common Mistakes
The most common error is to forget the coefficient \(\frac1a\), writing \(\operatorname{arcsec}\frac xa\) alone. Another is to confuse this integral with \(\int\frac{dx}{\sqrt{x^2-a^2}}\), which is logarithmic; the extra factor \(\frac1x\) is what changes the answer to an inverse trigonometric function. A third error is to drop the absolute value on the left branch, producing the wrong sign.
Quick Reference
| Integrand | Antiderivative |
|---|---|
| \(\frac{1}{x\sqrt{x^2-1}}\) | \(\operatorname{arcsec}|x|\) |
| \(\frac{1}{x\sqrt{x^2-9}}\) | \(\frac13\operatorname{arcsec}\frac{|x|}{3}\) |
| \(\frac{1}{x\sqrt{4x^2-1}}\) | \(\operatorname{arcsec}|2x|\) |
| \(\frac{1}{x\sqrt{x^2-a^2}}\) | \(\frac1a\arctan\frac{\sqrt{x^2-a^2}}{a}\ (x>a)\) |
Interpreting the Result
The antiderivative \(\frac1a\operatorname{arcsec}\frac xa\) increases from \(0\) at \(x=a\) toward the limit \(\frac{\pi}{2a}\) as \(x\to\infty\). The integrand decays like \(\frac{1}{x^2}\) for large \(x\), so the total area from \(a\) to infinity is finite: \(\int_a^\infty\frac{dx}{x\sqrt{x^2-a^2}}=\frac{\pi}{2a}\). Near \(x=a\) the integrand blows up like \(\frac{1}{a\sqrt{2a(x-a)}}\), which is integrable, so the improper integral is finite at both ends.
A Numerical Spot Check
For \(\int_{\sqrt2}^{2}\frac{dx}{x\sqrt{x^2-1}}=\frac\pi{12}\approx0.2618\), take the midpoint \(x=1.7071\), where the integrand is \(\frac{1}{1.7071\cdot1.3836}\approx0.4234\). The midpoint rule gives \(0.5858\cdot0.4234\approx0.248\), within six percent of the exact value, as expected from a single rectangle.
Practice Problems with Answers
Try each problem before reading the answer. Every answer below was checked by differentiation or by numerical integration.
- Evaluate \(\int_{\sqrt2}^{2}\frac{dx}{x\sqrt{x^2-1}}\). Hint: use \(\operatorname{arcsec}2=\frac\pi3\) and \(\operatorname{arcsec}\sqrt2=\frac\pi4\). Answer: \(\frac\pi{12}\)
- Evaluate \(\int_{2\sqrt2}^{4}\frac{dx}{x\sqrt{x^2-4}}\). Hint: include the factor \(\frac12\). Answer: \(\frac{\pi}{24}\)
- Find an antiderivative of \(\frac{1}{x\sqrt{x^2-9}}\) for \(x>3\). Hint: here \(a=3\). Answer: \(\frac13\operatorname{arcsec}\frac x3\)
- Evaluate \(\int_{-2}^{-\sqrt2}\frac{dx}{x\sqrt{x^2-1}}\). Hint: the integrand is negative on the left branch. Answer: \(-\frac\pi{12}\)
Worked Example: A Scaled Radical
Evaluate \(\int_{1/\sqrt2}^{1}\frac{dx}{x\sqrt{4x^2-1}}\).
Let \(u=2x\), so \(dx=\frac{du}{2}\) and \(x=\frac u2\). Then \(\frac{dx}{x\sqrt{4x^2-1}}=\frac{du/2}{(u/2)\sqrt{u^2-1}}=\frac{du}{u\sqrt{u^2-1}}\), so the antiderivative is \(\operatorname{arcsec}|2x|\). The factor \(\frac12\) from \(dx\) cancels the factor \(2\) from \(x=\frac u2\). At \(x=1\): \(\operatorname{arcsec}2=\frac\pi3\). At \(x=\frac{1}{\sqrt2}\): \(\operatorname{arcsec}\sqrt2=\frac\pi4\). Therefore
In this example the answer equals the first one, because the scaling \(x\mapsto2x\) leaves \(\frac{dx}{x}\) unchanged. That scale invariance is a useful check.
An Improper Integral to Infinity
Because \(\operatorname{arcsec}\frac xa\to\frac\pi2\) as \(x\to\infty\), the improper integral converges: \(\int_a^\infty\frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}\Bigl(\frac\pi2-0\Bigr)=\frac{\pi}{2a}\). For \(a=1\), this gives \(\frac\pi2\approx1.571\). The integral is improper at both ends: at \(x=a\) the integrand is unbounded but integrable, and at infinity it decays like \(x^{-2}\). Both pieces converge, and the antiderivative evaluates them directly through limits.
Relation to the Arccosine Form
The identity \(\operatorname{arcsec}u=\arccos\frac1u\) lets us write the antiderivative as \(\frac1a\arccos\frac{a}{|x|}\). In this form the argument \(\frac{a}{|x|}\) lies in \((0,1)\), which is the domain of arccosine, and the geometry is a triangle with hypotenuse \(|x|\) and adjacent side \(a\). Computer algebra systems often return this form, so it is good to recognize it. All three versions, secant, arccosine, and arctangent of the radical, differ by constants at most.
Why This Integral Is Not Logarithmic
The neighboring integral \(\int\frac{dx}{\sqrt{x^2-a^2}}\) led to \(\int\sec\theta\,d\theta\), a logarithm. Here the extra factor \(\frac1x=\frac{1}{a\sec\theta}\) exactly cancels the \(\sec\theta\) coming from \(dx\), so only \(\int d\theta\) is left. This is the same mechanism that gave \(\arcsin\) for \(\frac{1}{\sqrt{a^2-x^2}}\). The lesson is that the presence or absence of a single factor of \(x\) in the denominator decides whether the answer is an angle or a logarithm, and it is worth noticing before you start.
Domain and Continuity
The integrand is defined and continuous for \(|x|>a\) only. The two branches are separated by the gap \([-a,a]\), where the radicand is negative or zero, and an antiderivative may carry a different constant on each branch. A definite integral cannot cross the gap, so both endpoints must lie on the same side. For example, \(\int_{-3}^{3}\frac{dx}{x\sqrt{x^2-1}}\) is meaningless because the integrand is not real for \(|x|<1\).
When a problem gives endpoints on one branch, choose the formula matching that branch: \(\frac1a\operatorname{arcsec}\frac xa\) on the right and the same expression with \(|x|\) on the left. The absolute value version is safe on both, which is why it is the recommended form of the answer.
Where This Fits in the Course
This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Polynomial Factors Over √(x²−a²),” and the one that follows is “Reconstructing Variables With Right-Triangle Diagrams.” It completes the reciprocal forms for the radical of a difference of squares.
Check Your Understanding
Use the derivation and examples above to answer these questions.
- Why do the secant and tangent factors cancel after substituting \(x=a\sec\theta\)?
- Why is \(\sqrt{x^2-a^2}=-a\tan\theta\) on the left branch?
- Evaluate \(\int_{\sqrt2}^{2}\frac{dx}{x\sqrt{x^2-1}}\).
- Show that \(\frac1a\arctan\frac{\sqrt{x^2-a^2}}{a}\) is also an antiderivative for \(x>a\).
- Differentiate \(\operatorname{arcsec}x\) for \(x>1\) and confirm the integrand.
