Absolute Values in Logarithmic Back-Substitution

Why Absolute Values Matter in Back-Substitution

A secant substitution can turn a radical integral into an integral involving \(\sec\theta\). That trigonometric integral has a logarithmic antiderivative. The final step is to express the logarithm in terms of \(x\), while keeping track of which expressions are positive on the domain being considered.

This sign check matters because the real logarithm \(\ln u\) is defined only for \(u>0\). By contrast, \(\ln|u|\) is defined whenever \(u\ne 0\). A correct trigonometric antiderivative can therefore become incorrect if its absolute-value bars are dropped without checking the branch of the substitution.

Where the Logarithm Comes From

The relevant trigonometric antiderivative is that of \(\sec\theta\). Wherever \(\sec\theta\) and \(\tan\theta\) are defined, and \(\sec\theta+\tan\theta\ne 0\), differentiate using the chain rule:

The last equality follows by factoring the numerator as \(\sec\theta(\tan\theta+\sec\theta)\) and cancelling the nonzero denominator. Thus \(\ln|\sec\theta+\tan\theta|\) is an antiderivative of \(\sec\theta\) on any interval where the expression is defined. The absolute value makes the logarithm valid whether \(\sec\theta+\tan\theta\) is positive or negative.

Now consider the substitution \(x=a\sec\theta\) on the branch \(x>a\), with \(0\le\theta<\pi/2\). On this branch, \(\tan\theta\ge 0\), so

Also, \(dx=a\sec\theta\tan\theta\,d\theta\). Therefore the radical cancels in the integral:

Since \(\sec\theta=x/a\) and \(\tan\theta=\sqrt{x^2-a^2}/a\) on this branch, the logarithm becomes

Because \(a\) is a fixed positive constant, \(-\ln a\) can be absorbed into the arbitrary constant of integration. The resulting formula is \(\ln|x+\sqrt{x^2-a^2}|+C\). On \(x>a\), the quantity inside those absolute-value bars is in fact positive; retaining the bars is nevertheless useful for writing one formula that also works on the negative branch.

The Negative Branch Requires Its Own Sign Check

The substitution \(x=a\sec\theta\) with \(0\le\theta<\pi/2\) covers \(x\ge a\), not \(x\le -a\). To handle the negative branch, let \(x=-a\sec\theta\) with \(0\le\theta<\pi/2\). Then \(\sec\theta=-x/a>0\), and

Consequently, the integral becomes \(-\int\sec\theta\,d\theta\), and back-substitution gives

Here \(\sqrt{x^2-a^2}-x>0\) because \(x<-a<0\), so this logarithm needs no absolute-value bars. To compare it with the formula from the positive branch, use

For \(x<-a\), both factors on the left are negative. It follows that \(\left|x+\sqrt{x^2-a^2}\right|=a^2/(\sqrt{x^2-a^2}-x)\). Taking logarithms shows that the two branch formulas differ only by a constant. Thus one antiderivative formula works on both connected parts of the domain.

Worked Examples

A Common Mistake: Dropping Bars Without Checking the Domain

On the negative branch, \(x+\sqrt{x^2-a^2}<0\). Therefore \(\ln(x+\sqrt{x^2-a^2})\) is not a real-valued expression there. For example, at \(x=-2\) and \(a=1\), the quantity inside the logarithm is \(-2+\sqrt{3}<0\).

A tempting but invalid step is to write

The equality is false: the right-hand side is not defined as a real logarithm on that interval. The correction is to keep the absolute value, or to use the positive branch-specific expression from the substitution, \(-\ln((\sqrt{x^2-a^2}-x)/a)\). Absolute values are not decorative; they specify the real logarithm’s argument.

How to Choose the Correct Form

The same antiderivative can have different-looking expressions depending on the branch. The following summary separates the sign facts from the formula:

The expression \(x+\sqrt{x^2-a^2}\) never equals zero on either interval. If it were zero, then \(\sqrt{x^2-a^2}=-x\); squaring would give \(x^2-a^2=x^2\), contradicting \(a>0\). Thus the absolute-value logarithm is defined throughout both intervals.

Why the Branch Check Matters

Trigonometric substitution is not only a way to simplify a radical. It also selects a range of angles, and that range determines the signs of the trigonometric expressions. On the positive branch, \(x=a\sec\theta\) with \(0\le\theta<\pi/2\) makes both \(\sec\theta\) and \(\tan\theta\) nonnegative. On the negative branch, \(x=-a\sec\theta\) on the same angle range makes \(x\) negative while the radical remains positive.

Keeping these branches distinct prevents sign errors in \(dx\), in the radical, and in the logarithm. It also explains why a compact formula with absolute-value bars can be correct on both intervals even though a particular substitution may cover only one of them. As always, the intervals are disconnected, so the constants of integration on the two intervals need not be equal.