Detecting Cyclic Products Before Integrating

When Integration by Parts Comes Back to Its Starting Point

Integration by parts is especially useful when differentiating one factor makes it simpler, while integrating the other factor is manageable. Sometimes, however, the process does not simplify the remaining integral. Instead, it produces another integral of a closely related form. A second application may then bring back the original integral.

This return is not a failure. It creates an equation involving the integral we want. If the coefficient of that integral allows us to isolate it, the cycle gives an efficient solution.

Products of exponentials and sine or cosine are common examples. Differentiating sine and cosine alternates between them, while integrating an exponential preserves its form. The factors therefore keep reappearing in a predictable pattern.

What Counts as a Cycle?

For this tutorial, a cyclic integration-by-parts pattern is a sequence of integrations by parts in which the remaining integral returns to the starting integral, possibly with a coefficient, plus terms that can be integrated or evaluated directly.

For example, let \(I\) denote an antiderivative of \(e^x\sin x\), and let \(J\) denote an antiderivative of \(e^x\cos x\). The first integration by parts relates \(I\) to \(J\); the next relates \(J\) back to \(I\). The two equations can then be combined to find \(I\).

For indefinite integrals, equations between antiderivatives are understood up to an additive constant. In the calculations below, we choose particular antiderivatives while deriving the formula, then include \(+C\) in the final answer to represent the full family.

A Complete Cycle: Exponential Times Sine

Consider \(I=\int e^x\sin x\,dx\). Choose \(u=\sin x\) and \(dv=e^x\,dx\). Then \(du=\cos x\,dx\) and \(v=e^x\), so integration by parts gives

Set \(J=\int e^x\cos x\,dx\). For \(J\), choose \(u=\cos x\) and \(dv=e^x\,dx\). Then \(du=-\sin x\,dx\) and \(v=e^x\), giving

The last integral is \(I\). Substitute the expression for \(J\) into the equation for \(I\):

Now the original integral appears on both sides. Add \(I\) to each side and divide by \(2\):

The cycle made the problem solvable because collecting the two copies of \(I\) gave the nonzero coefficient \(2\).

Why the Cycle Produces an Equation

The method rests on the integration-by-parts identity, which follows from the product rule. If \(u\) and \(v\) are differentiable, then \((uv)'=u'v+uv'\). Integrating both sides gives \(\int u\,dv=uv-\int v\,du\), with the usual understanding that indefinite antiderivatives are determined up to constants.

In a cycle, this identity is applied to related products. Each application is valid under the usual conditions for integration by parts on the interval in question. Because the differentiated and integrated factors repeat, the resulting equations contain the same unknown integral. Substitution or rearrangement then reduces the cycle to an algebraic equation.

A General Exponential–Trigonometric Formula

The same cycle works for \(e^{ax}\sin(bx)\) and \(e^{ax}\cos(bx)\), where \(a\) and \(b\) are constants and \(a\ne 0\). Define

In the first integral, take \(u=\sin(bx)\) and \(dv=e^{ax}\,dx\). Then \(du=b\cos(bx)\,dx\) and \(v=e^{ax}/a\). In the second, take \(u=\cos(bx)\) and \(dv=e^{ax}\,dx\). These choices give the pair of equations

Substituting the second equation into the first yields

Since \(1+b^2/a^2=(a^2+b^2)/a^2\), and \(a^2+b^2>0\) when \(a\ne0\), the coefficient is nonzero. Solving gives the formulas below. The cosine formula follows by substituting the expression for \(I_s\) into the equation for \(I_c\).

The derivation above assumes \(a\ne0\), because it divides by \(a\). If \(a=0\), the integrands reduce to sine or cosine alone, and their elementary antiderivatives can be found directly. The displayed formulas also give the correct expressions when \(a=0\) and \(b\ne0\), but that case does not require a cycle.

When the Cycle Does Not Give a Solution

Not every return to the original integral is useful. Suppose algebra gives \(I=A+\lambda I\), where \(A\) is an expression already known. Rearranging gives \((1-\lambda)I=A\). The cycle determines \(I\) only if \(1-\lambda\ne0\). If \(1-\lambda=0\), the equation either reduces to an identity or signals an inconsistency in the calculation; it does not justify dividing by zero or conclude that the integral has no antiderivative.

In the exponential–sine cycle above, substitution gives a nonzero coefficient because \(a^2+b^2>0\) under the stated condition \(a\ne0\). A useful cycle is therefore more than a visual return: it must produce an equation that can actually be isolated.

A Common Sign Error

The sign in the second integration by parts is easy to lose. For \(J=\int e^x\cos x\,dx\), using \(u=\cos x\) gives \(du=-\sin x\,dx\). Thus

A common incorrect calculation drops the negative sign in \(du\) and writes \(J=e^x\cos x-I\). Substituting that wrong equation would give \(I=e^x\sin x-e^x\cos x+I\), which cancels \(I\) from both sides and leaves the false claim \(0=e^x(\sin x-\cos x)\). The contradiction is a warning to check the sign of \(du\), not evidence that the method failed. Correcting the sign gives \(J=e^x\cos x+I\), and then \(2I=e^x(\sin x-\cos x)\).

Recognizing When to Try This Method

Look for a product in which one factor has a derivative cycle and the other remains manageable when integrated. Exponentials paired with sine or cosine are a strong signal: the exponential retains its form, and the trigonometric factor alternates between sine and cosine. If the first application creates another integral of the same family, name it and continue only far enough to see whether the original integral returns.

By contrast, when differentiating a factor makes it progressively simpler, repeated integration by parts may reduce the problem rather than create a cycle. The choice of strategy depends on the behavior of the factors and on the algebraic equation the cycle produces. In either case, differentiation of the final antiderivative is a direct check on the result.