Checking Trigonometric-Substitution Antiderivatives

Why Check the Final Formula?

A trigonometric substitution can turn a radical integral into a trigonometric one, but the final answer must still be a function of the original variable \(x\). During the return to \(x\), it is possible to lose a chain-rule factor, choose an incorrect sign, or use a formula outside the interval where it is defined. Differentiating the proposed answer in \(x\) checks the result directly against the original integrand.

This is a useful final step, not a replacement for keeping track of the substitution. A derivative check establishes that a proposed expression is an antiderivative on the interval where the expression and the original integrand are defined. It does not make a formula valid at points where a radical or inverse trigonometric function is undefined.

For radical integrals, the interval matters. For example, \(\sqrt{a^2-x^2}\) is real when \(-a\le x\le a\), while its reciprocal is defined only when \(-a<x<a\). The expression \(\sqrt{x^2-a^2}\) is real when \(x\le -a\) or \(x\ge a\); its reciprocal is defined only on the corresponding open intervals. Here \(a\) is assumed positive.

A Reliable Verification Procedure

After back-substituting, set aside the trigonometric variable and differentiate the final formula as written in \(x\). Use the product rule for products involving a radical, the chain rule for inverse trigonometric functions, and the chain rule for logarithms. Then simplify until the derivative is visibly the original integrand.

Why Differentiating Is the Right Check

The method works because differentiation reverses integration: an antiderivative is defined by having derivative equal to the integrand. The chain rule also explains why the substitution steps must be tracked carefully. If the final expression has the form \(F(x)=G(u(x))\), then

The factor \(u'(x)\) is part of the derivative of the final expression. In a substitution-based calculation, the differential factor in the transformed integral accounts for this same change of variable. Once the answer has been returned to \(x\), differentiating it checks that the factor and the algebra have both been handled correctly.

More explicitly, suppose a proposed formula \(F\) is differentiable on an interval \(I\), and direct differentiation gives \(F'(x)=f(x)\) for all \(x\in I\). By the definition above, \(F\) is an antiderivative of \(f\) on \(I\). The check is complete only after the equality is established on the stated interval, not merely at a few sample values.

Worked Checks for \(a^2-x^2\)

Worked Check for \(x^2-a^2\)

Common Pitfall: Losing a Sign

A frequent error is to differentiate \(\sqrt{a^2-x^2}\) as though its derivative were \(x/\sqrt{a^2-x^2}\). That is the derivative pattern for \(\sqrt{x^2-a^2}\) on \(x>a\), not for \(\sqrt{a^2-x^2}\). The inner derivative in the first radical is \(-2x\), so its derivative has a minus sign.

For example, with \(S(x)=\sqrt{9-x^2}\), the incorrect calculation would give \(S'(x)=x/S(x)\). At \(x=1\), that predicts \(S'(1)=1/\sqrt{8}\). The correct chain-rule calculation is

The sign error changes the derivative of any product containing \(S(x)\), and the terms will not simplify to the original integrand. When checking an answer, compute each inner derivative explicitly before combining terms.

Domain and Comparison Checks

A successful derivative calculation must satisfy two conditions. First, the proposed expression must be defined and differentiable on the interval under consideration. Second, its derivative must equal the original integrand there. In particular, endpoints where a denominator becomes zero cannot be included in the interval for an improper-looking integrand such as \(1/\sqrt{a^2-x^2}\).

It is also important to compare with the original integrand, not merely with an expression that appeared in the trigonometric variable. For a radical, the back-substitution must produce the intended square root on the chosen domain. A triangle relation can give a positive length such as \(\sqrt{a^2-x^2}\); the domain of \(x\) determines which signs and inverse-trigonometric branches are appropriate when writing the answer in \(x\).