Why the Bounds and Branch Matter
For a definite integral, a substitution changes more than the integrand: it changes the variable and the limits as well. With \(\sqrt{x^2-a^2}\), there is an additional issue. For real \(x\), the radical is defined only when \(x^2-a^2\ge 0\), so the integration interval must lie on one of the two branches \(x\ge a\) or \(x\le -a\), where \(a>0\).
The secant substitution turns the expression under the radical into a trigonometric identity. To use it reliably in a definite integral, choose a branch, transform both endpoints, and preserve the order of the limits. If an endpoint equals \(a\) or \(-a\), the reciprocal radical is unbounded there; the integral must be interpreted as an improper integral and its limiting value checked.
Secant Substitution on the Positive Branch
When \(x\ge a\), use \(x=a\sec\theta\) with \(0\le\theta<\pi/2\). Then \(\sec\theta\ge1\) and \(\tan\theta\ge0\), so the principal square root is \(a\tan\theta\), not its negative. Differentiation gives \(dx=a\sec\theta\tan\theta\,d\theta\). For an endpoint \(x=c\ge a\), the corresponding angle is \(\theta=\operatorname{arcsec}(c/a)=\arccos(a/c)\).
In a definite integral from \(u\) to \(v\), where \(a\le u\le v\), the transformed limits are \(\theta_u=\arccos(a/u)\) and \(\theta_v=\arccos(a/v)\). They remain in increasing order because \(\arccos(a/x)\) increases as \(x\) increases on this branch. In particular, the endpoint \(x=a\) transforms to \(\theta=0\).
What the Substitution Does to the Two Radicals
The square-root factor and its reciprocal produce different trigonometric integrands. The identities below follow by inserting the substitution and cancelling common factors. They are useful starting points for definite integrals; the limits must still be transformed separately.
For the first expression, \(\tan^2\theta=\sec^2\theta-1\), so its integral reduces to a combination of \(\int\sec^3\theta\,d\theta\) and \(\int\sec\theta\,d\theta\). For the reciprocal, it reduces directly to \(\int\sec\theta\,d\theta\). The standard antiderivatives needed here are \(\frac12(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)\) for \(\sec^3\theta\), and \(\ln|\sec\theta+\tan\theta|\) for \(\sec\theta\). Their derivatives verify these formulas on \(0\le\theta<\pi/2\).
A Definite-Integral Formula and Its Justification
The substitution provides a direct way to calculate, but it is also useful to have endpoint formulas. Define, on either real branch \(x\ge a\) or \(x\le-a\),
For the reciprocal radical, define
The logarithm arguments are nonzero on both branches. The scale \(a\) makes the logarithm’s argument dimensionless; it does not change the derivative. These expressions give the following definite-integral rules when the endpoints lie on the same branch and the integrals exist:
Here is why the formulas work. Write \(R=\sqrt{x^2-a^2}\) at an interior point of either branch. Then \(R'=x/R\). By the product rule,
Also, differentiating the logarithm gives
Consequently,
The Fundamental Theorem of Calculus gives the endpoint formulas on intervals where the integrands are continuous. If an endpoint is \(a\) or \(-a\), apply the formulas by taking a one-sided limit from within the interval. The square-root integrand extends continuously to zero there. The reciprocal is unbounded, but its limiting integral is finite: in the transformed variable, it is an integral of \(\sec\theta\) starting or ending at \(\theta=0\), where \(\sec\theta\) is continuous.
Worked Examples
Worked Example: A Square-Root Integral from the Branch Endpoint
Evaluate \(\int_3^5\sqrt{x^2-9}\,dx\). Here \(a=3\), and the interval is on the positive branch. Set \(x=3\sec\theta\). The endpoint \(x=3\) gives \(\theta=0\), while \(x=5\) gives \(\sec\theta=5/3\), hence \(\tan\theta=4/3\). The transformed integral is
Using the antiderivatives above, an antiderivative of the transformed integrand is \(\frac92(\sec\theta\tan\theta-\ln|\sec\theta+\tan\theta|)\). At the upper limit, \(\sec\theta\tan\theta=(5/3)(4/3)=20/9\) and \(\sec\theta+\tan\theta=3\). At the lower limit, \(\sec 0=1\), \(\tan 0=0\), and \(\ln 1=0\). Therefore,
Check with \(F\): at \(x=5\), the radical is \(4\), so \(F(5)=\frac12(20-9\ln3)\); at \(x=3\), the radical is \(0\) and the logarithm is \(\ln1=0\), so \(F(3)=0\). Their difference is the same result.
Worked Example: A Reciprocal Radical at the Endpoint
Evaluate the improper integral \(\int_3^5\frac{dx}{\sqrt{x^2-9}}\). The integrand is unbounded at \(x=3\), so first interpret the integral as a limit:
Use \(x=3\sec\theta\). The lower limit \(b\) corresponds to \(\theta_b=\arccos(3/b)\), which tends to \(0\) as \(b\to3^+\). The upper limit is \(\alpha=\arccos(3/5)\), with \(\sec\alpha=5/3\) and \(\tan\alpha=4/3\). Since the transformed integrand is \(\sec\theta\),
The limit is finite, so the improper integral converges. As a check, \(G(5)=\ln((5+4)/3)=\ln3\), while \(G(3)=\ln(3/3)=0\). The endpoint difference is again \(\ln3\).
Worked Example: A Square-Root Integral on the Negative Branch
Evaluate \(\int_{-5}^{-3}\sqrt{x^2-9}\,dx\). The interval lies on \(x\le-3\), so use \(x=-3\sec\theta\) with \(0\le\theta<\pi/2\). Then \(\sqrt{x^2-9}=3\tan\theta\) and \(dx=-3\sec\theta\tan\theta\,d\theta\). At \(x=-5\), \(\sec\theta=5/3\), so \(\theta=\alpha=\arccos(3/5)\); at \(x=-3\), \(\theta=0\). Notice the transformed bounds decrease. Substitution gives
The last integral is the one evaluated in the first example: its value is \(10-\frac92\ln3\). The sign change comes from reversing the transformed limits, not from changing the sign of the square root. Check using \(F\): \(F(-3)=0\), and \(F(-5)=\frac12(-20-9\ln(1/3))=-10+\frac92\ln3\). Thus \(F(-3)-F(-5)=10-\frac92\ln3\), as required.
Endpoint and Branch Pitfalls
A common error is to use the positive-branch substitution \(x=a\sec\theta\) for a negative interval without accounting for the sign of \(x\) and the order of the transformed limits. For example, on \([-5,-3]\), writing \(x=3\sec\theta\) cannot represent any point in the interval: \(3\sec\theta\ge3\), whereas every \(x\) in the interval is negative. Even with the correct negative-branch substitution \(x=-3\sec\theta\), the bounds go from \(\alpha\) down to \(0\), not from \(0\) up to \(\alpha\). Reversing those bounds without also changing the sign would give a negative answer for an integral of a nonnegative function.
At a branch endpoint, do not substitute into an expression as if the reciprocal radical were defined there. Instead, use a one-sided limit. In this case the improper integral converges because the transformed integrand \(\sec\theta\) stays bounded as \(\theta\) approaches \(0\). This endpoint argument applies to a finite interval on either branch; it does not assert convergence for an interval extending to infinity.
Check Your Understanding
Use the branch, the transformed bounds, and endpoint behavior to guide each calculation.
- For an integral from \(a\) to \(c\), where \(c>a>0\), what are the secant-substitution limits?
- Why is \(\sqrt{x^2-a^2}=a\tan\theta\), rather than \(-a\tan\theta\), on the positive branch with \(0\le\theta<\pi/2\)?
- What substitution and transformed bounds apply on an interval \([u,v]\) with \(u<v\le-a\)?
- Why does the reciprocal radical have a finite improper integral at \(x=a\) over a finite interval on the positive branch?
- Use the endpoint formula for \(G\) to evaluate \(\int_4^5\frac{dx}{\sqrt{x^2-16}}\), interpreting the lower endpoint as improper.
