Repeated and Cyclic Integration by Parts

Overview

Central question: When does repeated integration by parts terminate or cycle? You will recognize two different patterns. In one, repeated differentiation eventually turns a polynomial into zero. In the other, repeated differentiation and integration return to the original integral, which can then be solved algebraically. These patterns look similar at first because both use the same rule, but their stopping logic is different.

Notice the factor $x^2$ in $\int x^2\cos(x)\,dx$. Each derivative makes it simpler: $x^2$, then $2x$, then $2$, then $0$. Predict what that sequence means before calculating. If one factor will disappear after a few derivatives, repeated parts has a natural endpoint.

Reduction to zero

Choose $u=x^2$ and $dv=\cos(x)\,dx$. Then $du=2x\,dx$ and $v=\sin(x)$, giving

$$\int x^2\cos(x)\,dx =x^2\sin(x)-\int2x\sin(x)\,dx.$$

Apply the rule again to the remaining integral, now with $u=2x$ and $dv=\sin(x)\,dx$. Because an antiderivative of $\sin(x)$ is $-\cos(x)$,

$$\int2x\sin(x)\,dx =-2x\cos(x)+\int2\cos(x)\,dx.$$

Combining the terms gives

$$\int x^2\cos(x)\,dx =x^2\sin(x)+2x\cos(x)-2\sin(x)+C.$$

The tabular method is a compact way to organize this repeated pattern. It does not replace reasoning. Before using it, verify that the differentiated column will end at zero and that repeated integration on the other side remains manageable. If either condition fails, a table can hide a poor choice rather than repair it.

When the integral comes back

Now consider

$$I=\int e^x\sin(x)\,dx.$$

Choose $u=\sin(x)$ and $dv=e^x\,dx$. One application gives

$$I=e^x\sin(x)-\int e^x\cos(x)\,dx.$$

Call the remaining integral $J$. Applying integration by parts to $J$ with $u=\cos(x)$ and $dv=e^x\,dx$ gives

$$J=e^x\cos(x)+\int e^x\sin(x)\,dx =e^x\cos(x)+I.$$

Substitute this result into the expression for $I$:

$$I=e^x\sin(x)-e^x\cos(x)-I.$$

The original integral has returned. Add $I$ to both sides, then divide by two:

$$I=\frac{e^x}{2}\left(\sin(x)-\cos(x)\right)+C.$$

Do not confuse recurrence with failure

The reappearance of $I$ is not an error. It becomes useful only when the returned copy has a coefficient that can be moved to the other side. If the original integral reappears unchanged in an equation such as $I=I+$ a nonzero expression, check the signs and antiderivatives; that would be contradictory. If it appears as $I=A-I$, the cycle has created solvable algebra.

Check the sign pattern by differentiation

The compact form of repeated parts can make a sign error look respectable. Verify the terminating example by differentiating

$$x^2\sin(x)+2x\cos(x)-2\sin(x).$$

The derivative of the first term is $2x\sin(x)+x^2\cos(x)$. The derivative of the second is $2\cos(x)-2x\sin(x)$. The derivative of the third is $-2\cos(x)$. Everything except $x^2\cos(x)$ cancels. This is the desired integrand, and the cancellation explains why the alternating signs in the tabular arrangement were necessary.

The cyclic result can be checked the same way. Differentiate

$$\frac{e^x}{2}\left(\sin(x)-\cos(x)\right).$$

Use the product rule. Differentiating $e^x$ reproduces the bracket, while differentiating the bracket gives $\cos(x)+\sin(x)$. Adding them cancels the negative cosine and doubles sine, leaving $e^x\sin(x)$. A check is especially important in cyclic work because one wrong sign may not become obvious until the final algebraic isolation.

A table is an abbreviation, not an argument

For a polynomial times an exponential or trigonometric factor, a table can record repeated derivatives and antiderivatives efficiently. It should never be the first line on paper. First identify the factor that terminates under differentiation and the factor whose antiderivatives stay controlled. Then make one full application of integration by parts in ordinary notation. Only after the relationship is clear should a table replace repeated lines.

This policy also tells you when not to use the tabular shortcut. It is poor for $\int\ln(x)\sin(x)\,dx$, because differentiating $\ln(x)$ never reaches zero and produces increasingly complicated negative powers of $x$. The table may be visually neat while the mathematics is getting worse. Choose a representation because it reveals a pattern, not because it reduces the amount of writing.

Stop and check

A common wrong turn is to keep applying parts to $\int e^x\sin(x)\,dx$ after $I$ returns, hoping the expression will eventually end. The misconception is assuming every repeated method must terminate. The targeted correction is to name the returned integral, substitute it into the first equation, and solve the equation for that named quantity.

Retrieval check

Explain which pattern applies to $\int x^3e^x\,dx$ and which applies to $\int e^x\cos(x)\,dx$. In the first, repeated differentiation of the polynomial reaches zero. In the second, the sine-cosine derivative cycle returns the original form. Differentiate any final antiderivative to verify the sign pattern.

The next lesson changes the structural signal. Instead of a product-rule reversal, powers of sine and cosine invite identity choices that expose a usable derivative factor.