Definite Integrals With a^2 + x^2 Radicals

Definite Integrals Under the Tangent Substitution

The tangent substitution \(x=a\tan\theta\) is especially clean for definite integrals of \(\sqrt{a^2+x^2}\) and \(\frac{1}{\sqrt{a^2+x^2}}\), because the new limits are given by \(\theta=\arctan\frac xa\), and for many familiar limits this is a standard angle such as \(\frac\pi4\) or \(\frac\pi3\). Evaluating at those angles avoids back-substituting a radical and makes the final arithmetic exact.

How the Limits Transform

With \(x=a\tan\theta\), we have \(\sqrt{a^2+x^2}=a\sec\theta\) and \(dx=a\sec^2\theta\,d\theta\). Then

where \(\theta_i=\arctan\frac{x_i}{a}\). Because \(\tan\) maps \((-\frac\pi2,\frac\pi2)\) one-to-one onto the real line, every real limit has exactly one angle, and the integrals never meet a singularity. This is a significant simplification compared with the secant and sine substitutions, which restrict \(x\) to part of the real line. The antiderivatives are \(\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|\) and \(\int\sec^3\theta\,d\theta=\frac12(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)\).

Worked Examples

Application: Arc Length of a Parabola

For the parabola \(y=\frac{x^2}{2}\), the arc length element is \(\sqrt{1+y'^2}\,dx=\sqrt{1+x^2}\,dx\). The length from \(x=0\) to \(x=1\) is therefore \(\int_0^1\sqrt{1+x^2}\,dx=\frac12\bigl[\sqrt2+\ln(1+\sqrt2)\bigr]\approx1.148\). This is slightly longer than the straight chord from \((0,0)\) to \((1,\frac12)\), which has length \(\sqrt{1.25}\approx1.118\), as an arc must be.

A Reliable Procedure

Common Mistakes

The most common error is to transform the integrand but leave the limits in terms of \(x\), producing nonsense. Another is to forget the factor \(a^2\) in front of the \(\sec^3\) integral. A third is to evaluate \(\ln(x+\sqrt{a^2+x^2})\) at \(x=0\) as zero when \(a\ne1\), which ignores the constant \(\ln a\). A fourth is to treat the domain as restricted, when in fact the radical is defined for all real \(x\).

A Numerical Spot Check

Test \(\int_0^1\frac{dx}{\sqrt{1+x^2}}=0.881\) with Simpson's rule on three points: the heights are \(1\), \(\frac{1}{\sqrt{1.25}}\approx0.8944\), and \(\frac{1}{\sqrt2}\approx0.7071\). Then \(\frac{0.5}{3}(1+4\cdot0.8944+0.7071)\approx0.8808\), which agrees with the exact value to three digits.

Practice Problems with Answers

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

1. Evaluate \(\int_0^1\frac{dx}{\sqrt{1+x^2}}\). Hint: the angle limits are \(0\) and \(\frac\pi4\). Answer: \(\ln(1+\sqrt2)\)
2. Evaluate \(\int_0^{\sqrt3}\sqrt{1+x^2}\,dx\). Hint: use \(\frac12[\sec\theta\tan\theta+\ln(\sec\theta+\tan\theta)]\) at \(\frac\pi3\). Answer: \(\sqrt3+\frac12\ln(2+\sqrt3)\)
3. Evaluate \(\int_0^2\sqrt{4+x^2}\,dx\). Hint: include the factor \(a^2=4\). Answer: \(2\sqrt2+2\ln(1+\sqrt2)\)
4. Evaluate \(\int_{-1}^{2}\frac{dx}{\sqrt{1+x^2}}\). Hint: use \(F(x)=\ln(x+\sqrt{1+x^2})\). Answer: \(\ln\bigl[(2+\sqrt5)(\sqrt2+1)\bigr]\)

Why the Closed Form and the Angle Form Agree

The two ways of evaluating each integral are the same calculation expressed in different variables. In the angle form, one evaluates \(\sec\theta\) and \(\tan\theta\) at the transformed limits. In the closed form, one evaluates \(x\) and \(\sqrt{a^2+x^2}\) at the original limits. The relations \(\tan\theta=\frac xa\) and \(\sec\theta=\frac{\sqrt{a^2+x^2}}{a}\) connect the two. Because \(\frac{\sec\theta+\tan\theta}{1}=\frac{x+\sqrt{a^2+x^2}}{a}\), the logarithms differ only by the constant \(\ln a\), which cancels in any definite integral. This cancellation is why the lower limit \(x=0\) contributes \(\ln a\) in the closed form even though the angle form contributes zero.

Checking the Original Domain

A final step in any substitution problem is to confirm that the transformed integral is equivalent to the original one. Here the check is straightforward. The map \(x=a\tan\theta\) is continuous and strictly increasing from \((-\frac\pi2,\frac\pi2)\) onto \((-\infty,\infty)\), and \(\sqrt{a^2+x^2}=a\sec\theta\) holds because \(\sec\theta>0\) on this range. The integrand is continuous everywhere, so no limit is improper, and no singularity is crossed. For contrast, the sine substitution requires \(|x|\le a\), and the secant substitution requires \(|x|\ge a\), and both need explicit domain checks.

Interpreting the Arc-Length Application

The value \(\frac12[\sqrt2+\ln(1+\sqrt2)]\approx1.148\) for the arc of the parabola between \(x=0\) and \(x=1\) is compared with two bounds. The chord from \((0,0)\) to \((1,\frac12)\) has length \(\sqrt{1.25}\approx1.118\), and the path along the horizontal and vertical legs of the corner has length \(1.5\). The arc length lies between them, as it must. The formula also shows how a transcendental number such as \(\ln(1+\sqrt2)\) arises naturally from a simple geometric curve.

Large-Interval Behavior

For a long interval \([0,b]\) with large \(b\), the integral \(\int_0^b\frac{dx}{\sqrt{a^2+x^2}}\approx\ln\frac{2b}{a}\) grows only logarithmically, while \(\int_0^b\sqrt{a^2+x^2}\,dx\approx\frac{b^2}{2}+\frac{a^2}{2}\ln\frac{2b}{a}+\frac{a^2}{4}\) grows quadratically. These estimates come from the closed forms and are useful for sanity-checking results with large limits.

Where This Fits in the Course

This tutorial belongs to the module “Trigonometric Substitution”. The tutorial just before it is “Integrals With x√(a^2 + x^2) in the Denominator,” and the one that follows is “Proper Rational Functions And Partial Fraction Forms.” It brings the tangent substitution to definite integrals, where the transformed limits make the arithmetic exact. It also closes the group of trigonometric substitutions in this module, so a good next step is to practice choosing among the sine, tangent, and secant substitutions from the radicand alone, before moving on to partial fractions and completing the square.