When Integration by Parts Cycles
Some integrals become simpler after integration by parts because one factor eventually disappears. For instance, repeated integration by parts works well when differentiating a polynomial: its degree falls at each step. Exponential-sine and exponential-cosine products behave differently. Differentiating sine produces cosine, and differentiating cosine produces sine. Integrating an exponential leaves an exponential factor. The same kinds of terms therefore reappear.
This repetition is not a failure of the method. It is a signal that the integrals should be treated as a pair. One integration-by-parts step relates the sine integral to the cosine integral; a second step relates the cosine integral back to the sine integral. Keeping both equations visible makes the cycle manageable.
Set Up a Matched Pair
Let \(a\) and \(b\) be real constants, and first suppose \(a\ne 0\). For a fixed base point \(x_0\), define the two accumulated integrals
Using a fixed lower endpoint avoids ambiguity about additive constants: \(S(x_0)=C(x_0)=0\), and each integral has a definite value for every \(x\). The dummy variable \(t\) is used inside the integrals so it is distinct from the endpoint \(x\).
For the sine integral, take \(u=\sin(bt)\) and \(dv=e^{at}\,dt\). Then \(du=b\cos(bt)\,dt\) and \(v=e^{at}/a\). For the cosine integral, take \(u=\cos(bt)\) and \(dv=e^{at}\,dt\). In the second calculation, the derivative \(du=-b\sin(bt)\,dt\) is responsible for a plus sign in the resulting relation.
These are the cyclic equations. Each equation contains one of the two integrals and a multiple of the other. The boundary terms depend on the chosen interval; the coefficients linking the integrals come from differentiating the trigonometric factors.
Why Two Steps Return to the Starting Integral
The cycle follows directly from the two derivatives \(\frac{d}{dt}\sin(bt)=b\cos(bt)\) and \(\frac{d}{dt}\cos(bt)=-b\sin(bt)\). The first integration-by-parts step changes sine to cosine. The second changes cosine back to sine. Substitution therefore returns to an equation involving the original \(S(x)\).
To see the algebra without dropping the endpoint terms, write
The two equations become \(S=P-\frac{b}{a}C\) and \(C=Q+\frac{b}{a}S\). Substitute the second into the first. Every step is an equality, and the coefficient of \(S\) comes from moving the returned \(S\)-term to the left.
This is the basic reason the cycle is useful: after two steps, the original integral has not proliferated into a longer chain. It satisfies an algebraic relation with explicit boundary terms. Since \(a\ne0\), the coefficient \(1+b^2/a^2\) is positive, so it is nonzero. The same substitution can be made in the other direction to obtain an equation for \(C\).
Worked Example: A Cycle with Equal Exponent and Frequency
Worked Example: A Cycle with Equal Exponent and Frequency
Use integration by parts to find \(S(x)=\int_0^x e^t\sin t\,dt\), keeping the paired cosine integral \(C(x)=\int_0^x e^t\cos t\,dt\) visible.
Here \(a=b=1\). The sine equation has boundary term \([e^t\sin t]_0^x=e^x\sin x\), since \(e^0\sin 0=0\). The cosine equation has boundary term \([e^t\cos t]_0^x=e^x\cos x-1\), since \(e^0\cos 0=1\). Thus
Substitute the second equation into the first:
Check by differentiating the result. The derivative of the numerator \(e^x(\sin x-\cos x)\) is \(e^x(\sin x-\cos x)+e^x(\cos x+\sin x)=2e^x\sin x\). Dividing by \(2\) gives \(S'(x)=e^x\sin x\), as required. At \(x=0\), the expression is \((0-1+1)/2=0\), matching the defining lower endpoint.
Worked Example: Keep the Frequency Coefficients
Worked Example: Keep the Frequency Coefficients
Find \(C(x)=\int_0^x e^{2t}\cos(3t)\,dt\) by pairing it with \(S(x)=\int_0^x e^{2t}\sin(3t)\,dt\).
Now \(a=2\) and \(b=3\). Applying the two cyclic equations gives
The \(-1\) in the second boundary term is necessary because \(e^0\cos 0=1\). Substitute the expression for \(S\) into the equation for \(C\):
For the check, differentiate the numerator's \(x\)-dependent terms. Their derivative is \(2e^{2x}\cos(3x)-6e^{2x}\sin(3x)+6e^{2x}\sin(3x)+9e^{2x}\cos(3x)=13e^{2x}\cos(3x)\). Dividing by \(13\) gives \(C'(x)=e^{2x}\cos(3x)\). At \(x=0\), the numerator is \(2-2=0\), as required.
Worked Example: Organize a Definite Cyclic Integral
Worked Example: Organize a Definite Cyclic Integral
Evaluate \(A=\int_0^{\pi/2}e^t\sin t\,dt\) by setting up the paired integral \(B=\int_0^{\pi/2}e^t\cos t\,dt\). The endpoint values of sine and cosine make the boundary terms easy to track.
For the sine integral, \([e^t\sin t]_0^{\pi/2}=e^{\pi/2}-0\), so \(A=e^{\pi/2}-B\). For the cosine integral, \([e^t\cos t]_0^{\pi/2}=0-1=-1\), so \(B=-1+A\). Consequently,
Check this value using the antiderivative from the previous example, \(\frac{e^t(\sin t-\cos t)+1}{2}\). At \(t=\pi/2\), it is \((e^{\pi/2}+1)/2\); at \(t=0\), it is \(0\). Their difference matches the computed definite integral.
Signs, Special Cases, and Common Pitfalls
The cycle is short, so a single sign error can change the final coefficient or result. In the cosine calculation, the derivative of \(\cos(bt)\) is \(-b\sin(bt)\). Integration by parts also subtracts the integral of \(v\,du\). These two negative signs combine to give a plus sign in the equation for \(C\).
Two special cases are worth separating from the genuine cycle. If \(b=0\), then \(\sin(bt)=0\) and \(\cos(bt)=1\): the sine integral is zero, and the cosine integral is just an exponential integral. If \(a=0\), the exponential factor is \(1\), so the integrals are ordinary trigonometric integrals. The cyclic equations above should not be used with \(a=0\), because they divide by \(a\).
For a definite integral, do not discard endpoint terms. The cyclic equations include the values at both endpoints, and those values may be unequal even when one endpoint makes a sine or cosine term vanish. For an indefinite integral, integration by parts yields relations between antiderivatives up to additive constants. Definite integrals from a common base point avoid that bookkeeping issue.
A Reliable Way to Organize the Cycle
When an exponential-sine or exponential-cosine product appears, use a consistent setup rather than repeating parts informally:
Use one symbol for the sine product and another for the cosine product, with the same exponent, frequency, and limits.
If integrating the exponential as \(e^{at}/a\), confirm that \(a\ne0\).
Write the boundary term and the remaining paired integral, keeping the derivative sign of cosine explicit.
Collect the returned integral terms on one side; this exposes the cycle's coefficient.
For an antiderivative, differentiate it. For definite integrals, check the endpoint values and confirm the paired equations use the same limits.
This organization is useful because it turns a potentially confusing sequence of repeated integrations by parts into a small, auditable system. It also makes clear which features are structural—the return from sine to cosine and back—and which depend on the particular interval through the boundary terms.
Check Your Understanding
Use the paired-integral viewpoint to answer these questions.
- Why does integration by parts for an exponential times sine naturally introduce an exponential times cosine?
- For \(a\ne0\), what sign appears in front of the sine integral in the equation for the cosine integral, and why?
- What condition is required to use \(e^{at}/a\) as the antiderivative of \(e^{at}\)?
- When defining paired definite integrals, why should they use the same endpoints?
- What is one appropriate check for a computed antiderivative, and what endpoint check applies when the integral starts at \(x_0\)?
