Why Check an Antiderivative by Differentiating?
An antiderivative is defined by the derivative it produces. That makes differentiation the direct test of a proposed answer: differentiate the candidate and compare the result with the function being integrated. This check is especially useful for trigonometric functions, where a correct formula can differ from a plausible-looking one by a minus sign or by a different trigonometric function.
For instance, the derivatives of sine and cosine involve each other, and the derivatives of tangent and cotangent have opposite signs. Remembering only that a formula “looks familiar” is not enough. A reliable check asks whether the derivative is exactly the target function on the interval under consideration.
When an indefinite integral is written with \(+C\), the constant accounts for the fact that adding a constant does not change a derivative. The check is still the same: differentiate the expression, and verify that its derivative is the integrand. In the examples below, we work on intervals where all displayed functions are defined.
A Reliable Checking Method
Start by identifying the target function and the proposed antiderivative. Then differentiate the proposed expression carefully, using the ordinary derivative rules and the domain restrictions of its terms. Finally, compare the derivative with the target. If they agree, the candidate passes; if not, the derivative tells you what must be corrected.
State the function \(f(x)\) that the candidate is supposed to differentiate to.
Keep signs, function names, and any constants or logarithms visible in the calculation.
Check whether the derivative equals \(f(x)\), not merely a related trigonometric function.
Adjust the sign or choose a candidate whose derivative is the required function, then differentiate the corrected expression.
The basic derivative facts most often needed for this check are:
These formulas expose common mismatches. For example, differentiating \(\cot x\) gives \(-\csc^2 x\), not \(\sec^2 x\). Differentiating \(\cos x\) gives \(-\sin x\), not \(\sin x\). The target function determines which result is needed.
Worked Examples: Find the Error and Correct It
Worked Example: A Sign Error for Sine
Check the proposal \(F(x)=\cos x\) as an antiderivative of \(f(x)=\sin x\).
Differentiate the proposal:
The target is \(\sin x\), whereas the derivative is \(-\sin x\). Thus the proposed function has the wrong sign. To correct it, use \(G(x)=-\cos x+C\), where \(C\) is constant. Its derivative is
The corrected candidate passes: differentiating it gives the target \(\sin x\).
Worked Example: A Function Mismatch for Cosine
Check the proposal \(F(x)=-\sin x\) as an antiderivative of \(f(x)=\cos x\).
Differentiate each factor, keeping the negative sign:
This derivative is not \(\cos x\); it is its negative. The correction is \(G(x)=\sin x+C\). Differentiating verifies the correction:
Therefore \(F(x)=-\sin x\) fails for this target, while \(G(x)=\sin x+C\) is a valid antiderivative of \(\cos x\).
Worked Example: A Trigonometric Function Mismatch
Suppose the target is \(f(x)=\sec^2 x\), and the proposed candidate is \(F(x)=-\cot x\). Work on an interval where the expressions are defined.
Differentiate the proposal, including its negative sign:
The result is \(\csc^2 x\), not \(\sec^2 x\), so this is a function mismatch rather than a sign mismatch. A suitable correction is \(G(x)=\tan x+C\). Indeed,
Thus the corrected candidate differentiates to the required target.
Checking Logarithmic Trigonometric Antiderivatives
Some basic trigonometric antiderivatives involve logarithms. In these cases, a useful check is to apply the derivative rule \(\frac{d}{dx}\ln|u(x)|=\frac{u'(x)}{u(x)}\), wherever \(u(x)\ne 0\). The absolute value does not remove the need to check where the logarithm is defined; the expression inside it must be nonzero.
Worked Example: Correcting the Sign for Cosecant
On an interval where the expressions are defined, check \(F(x)=\ln|\csc x+\cot x|\) as an antiderivative of \(f(x)=\csc x\).
Set \(u(x)=\csc x+\cot x\). Its derivative is
Since \(u(x)=\csc x+\cot x\), the logarithm rule gives
The proposal has the wrong sign for the target \(\csc x\). Correct it by taking \(G(x)=-\ln|\csc x+\cot x|+C\). Then
The corrected expression passes the derivative test on any interval where it is defined.
The calculation also illustrates why the logarithm rule should be applied to the whole inner expression. Differentiating just one term of \(\csc x+\cot x\), or overlooking the negative signs in both derivative formulas, can lead to an incorrect conclusion.
Why the Constant Is the Only Difference on an Interval
If two differentiable functions have the same derivative on an interval, they differ by a constant there. This fact explains why an antiderivative check identifies a family of answers rather than a single expression. The proof uses the Mean Value Theorem: for any two points in the interval, the change in the difference of the functions is determined by its derivative at an intermediate point.
To prove this, define \(H(x)=F(x)-G(x)\). For every \(x\in I\),
Take any \(a,b\in I\) with \(a<b\). Since \(I\) is an interval, \([a,b]\) lies in \(I\). The function \(H\) is continuous on \([a,b]\) and differentiable on \((a,b)\), so the Mean Value Theorem gives a \(c\in(a,b)\) such that
Therefore \(H(b)=H(a)\) for any such pair of points. Hence \(H\) is constant on \(I\), which means \(F(x)=G(x)+C\) for some constant \(C\). Conversely, adding any constant to an antiderivative leaves its derivative unchanged, because \(\frac{d}{dx}C=0\).
Common Pitfalls
A frequent mistake is to remember the right trigonometric function but miss its sign. Another is to treat related functions as interchangeable: \(\sec^2 x\) and \(\csc^2 x\) are different, as are \(\sec x\tan x\) and \(\csc x\cot x\). The correction should follow from the derivative calculation rather than from a visual resemblance between formulas.
Also check constants and parentheses. For example, a constant added outside an expression contributes zero to the derivative, but a negative sign multiplying an entire expression changes the sign of its derivative. Write enough intermediate steps to make those effects visible.
Why This Check Matters
Differentiation turns an antiderivative question into a direct comparison that can settle whether a candidate is correct. It can distinguish a valid answer from one that is nearly correct, pinpoint a sign error, and reveal when the candidate produces the wrong trigonometric function. It also provides a final verification after an antiderivative has been found by another method.
For a proposed \(F\) and target \(f\), the essential question is simply whether \(F'(x)=f(x)\) throughout the stated interval. If the equality fails, inspect the differentiated expression: a negative of the target indicates a sign problem, while a different trigonometric function indicates a function mismatch. After correcting the candidate, differentiate it once more rather than assuming the adjustment worked.
Check Your Understanding
For each question, use differentiation to justify your conclusion and state any relevant domain restriction.
- Is \(\sin x+C\) an antiderivative of \(\cos x\)? Show the derivative.
- A candidate for an antiderivative of \(\sin x\) is \(\cos x+C\). What does differentiating it show, and how can it be corrected?
- Differentiate \(-\cot x\). Is the result \(\sec^2 x\) or \(\csc^2 x\)?
- Why does adding a constant to a valid antiderivative leave it valid?
- If two antiderivatives of the same function are defined on an interval, why can they differ only by a constant?
