Why Sign Checks Matter
Repeated integration by parts can produce a correct-looking answer that is wrong because one sign changed, a frequency factor disappeared, or an equation was rearranged incorrectly. These errors can be difficult to spot by scanning the final expression. A reliable check is to differentiate the proposed antiderivative and compare the result with the original integrand.
This tutorial focuses on diagnosing those errors in tabular integration and cyclic integration by parts. The methods themselves rely on integration by parts and the chain rule; the emphasis here is on tracking each sign and coefficient and checking the result.
Frequency Factors in Sine and Cosine Antiderivatives
The chain rule explains why integrating a trigonometric function with frequency \(k\) introduces a factor of \(1/k\), when \(k\ne 0\). For example, differentiating \(-\cos(kx)/k\) gives \(\sin(kx)\): the derivative of \(-\cos(kx)\) is \(k\sin(kx)\), and division by \(k\) cancels that factor.
Worked Example: Finding a Missing Frequency Factor
Flawed calculation. Suppose a calculation claims \(\int \sin(4x)\,dx=-\cos(4x)+C\). Differentiate the proposed answer:
The derivative is \(4\sin(4x)\), not the required \(\sin(4x)\). The proposed answer is too large by a factor of \(4\).
Correction. Divide by the frequency:
Check.
The sign and the factor are both necessary. The companion cosine formula gives \(\int\cos(4x)\,dx=\sin(4x)/4+C\); differentiating it likewise returns \(\cos(4x)\).
Why the Tabular Signs Alternate
In tabular integration, repeated integration by parts pairs successive derivatives of one factor with successive antiderivatives of the other. The signs alternate because each application of integration by parts introduces a subtraction.
More precisely, let \(P\) be a polynomial of degree \(n\), let \(f\) have a sequence of antiderivatives \(F_0,F_1,\ldots,F_n\), and define them by \(F_0'=f\) and \(F_{j+1}'=F_j\). The alternating expression in the following formula is an antiderivative of \(Pf\):
Here is a direct justification. Differentiate the sum. Differentiating each \(P^{(j)}\) produces a term involving \(P^{(j+1)}F_j\); differentiating \(F_j\) produces \(P^{(j)}F_{j-1}\) when \(j\ge 1\), since \(F_j'=F_{j-1}\). Those two kinds of terms cancel in adjacent positions because their signs are opposite. The first remaining term is \(P F_0'=Pf\), while the final derivative of the polynomial vanishes because \(P^{(n+1)}=0\). Thus the derivative of the sum is exactly \(Pf\), proving the formula.
Worked Example: Correcting a Tabular Sign
Evaluate \(\int x^2\sin(3x)\,dx\). First list the derivatives of the polynomial and successive antiderivatives of the trigonometric factor:
The antiderivatives are checked in order: the derivative of \(-\cos(3x)/3\) is \(\sin(3x)\); the derivative of \(-\sin(3x)/9\) is \(-\cos(3x)/3\); and the derivative of \(\cos(3x)/27\) is \(-\sin(3x)/9\). The table contributes with signs \(+,-,+\), so
Simplifying gives
Diagnosing a sign error. If the middle table sign is incorrectly taken as negative, the second term would be \(-2x\sin(3x)/9\). Differentiating that flawed expression gives
which contains extra terms and is not the integrand. For the corrected expression, differentiate each term:
The terms other than \(x^2\sin(3x)\) cancel, confirming the answer.
Diagnosing Cyclic Integration Errors
For an exponential multiplied by a sine or cosine, integration by parts can return to the other trigonometric integral. This creates a pair of equations. A sign error in either equation, or a lost coefficient when substituting one into the other, changes the result.
For \(a\ne 0\), define \(I\) and \(J\) as antiderivatives of \(e^{ax}\sin(bx)\) and \(e^{ax}\cos(bx)\), respectively. Applying integration by parts gives the coupled relations, with additive constants suppressed while solving for particular antiderivatives:
The minus sign in the first relation comes from differentiating \(\sin(bx)\), which gives \(b\cos(bx)\), in the integration-by-parts step. The plus sign in the second comes from differentiating \(\cos(bx)\), which gives \(-b\sin(bx)\), and subtracting the resulting integral.
Worked Example: Keeping the Coefficient When Solving a Cyclic Pair
Consider \(I=\int e^x\sin x\,dx\) and \(J=\int e^x\cos x\,dx\). The coupled relations are
Flawed rearrangement. Substituting the second equation into the first gives \(I=e^x\sin x-e^x\cos x-I\). A common error is to drop the final \(-I\) and conclude \(I=e^x(\sin x-\cos x)\). The coefficient must be retained when collecting the \(I\) terms.
Correction. Substitute, then add \(I\) to both sides:
Therefore a particular antiderivative is \(I=e^x(\sin x-\cos x)/2\). The flawed answer fails the derivative check:
It produces twice the required integrand. For the corrected answer,
Thus \(\int e^x\sin x\,dx=e^x(\sin x-\cos x)/2+C\).
A Second Check on Cyclic Signs and Frequencies
When the exponential and trigonometric factors have different rates, both the exponential parameter and the trigonometric frequency enter the answer. Solving the coupled relations for \(a\ne 0\) yields
provided \(a^2+b^2\ne 0\). The denominator and the sign before the cosine are worth checking rather than memorizing blindly.
Worked Example: Checking Both Parameters in a Cyclic Antiderivative
Evaluate \(\int e^{2x}\sin(3x)\,dx\). Here \(a=2\), \(b=3\), and \(a^2+b^2=4+9=13\). The formula gives
Differentiate using the product rule. The derivative of the exponential contributes \(2e^{2x}(2\sin(3x)-3\cos(3x))/13\). The derivative of the trigonometric factor is \(6\cos(3x)+9\sin(3x)\). Combining these contributions gives
The cosine terms cancel, and the sine coefficients add to \(13\), matching the denominator. This check detects a missing frequency factor or an incorrect sign in the numerator.
A Practical Error-Checking Routine
When reviewing a tabular or cyclic calculation, check the construction before relying on the final formula. The following sequence isolates the usual sources of error:
For trigonometric entries, differentiate each proposed antiderivative immediately. This reveals a wrong sign or a missing frequency divisor before it spreads through the table.
In a tabular sum, mark \(+,-,+,-,\ldots\) separately from the entries. Apply the sign to the whole product in that row.
Use parentheses around the entire expression being substituted, then collect all terms involving the unknown integral on one side.
Expand enough to see whether unwanted terms cancel and whether the remaining coefficient is exactly one.
A frequent mistake is to check only that the final expression has the expected general shape. A plausible exponential-trigonometric numerator is not sufficient: the coefficients must make its derivative equal the original integrand. Likewise, a table with alternating signs is not correct merely because the pattern looks familiar; each antiderivative entry must also be correct.
Check Your Understanding
Use differentiation to check proposed antiderivatives wherever possible.
- What factor is required in an antiderivative of \(\sin(5x)\), and why?
- For a tabular sum, what sign belongs to the third product, counting the first product as the first?
- Differentiate \(-\cos(2x)/2\). What does the result tell you about the frequency factor?
- In the cyclic equations for \(\int e^x\sin x\,dx\), why does substitution produce \(2I\) rather than \(I\) on the left?
- What should you compare after differentiating a proposed antiderivative?
