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A power of $x$ comes down one application at a time, the signs alternate, and an integral that returns to itself is finished by solving for it rather than by integrating again.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can apply integration by parts more than once, keeping the alternating sign that each application contributes, and say how many applications a given power of $x$ will need. You can lay a repeated calculation out so the signs cannot get lost, use the vanishing of a boundary term to shorten a definite integral, and recognise a circular integral as an equation to be solved rather than a calculation that has failed.
You can apply $\int u\,dv = uv - \int v\,du$ once, choosing $u$ to be the factor that simplifies when differentiated. Everything here is that same step done more than once — and the two things that go wrong when it is are both bookkeeping rather than calculus.
A repeated application: parts used again on the integral parts produced.
The differentiated column: the successive derivatives of $u$, ending at zero when $u$ is a polynomial.
The alternating sign: each application contributes one more minus, so the terms run $+,-,+,-$.
A circular integral: one where two applications return the original integral, so the answer is found by solving rather than by integrating.
A power of $x$ comes down one step at a time. In $\int x^{2}e^{x}dx$ the first application replaces $x^{2}$ by $2x$ and leaves $\int 2x e^{x}dx$, which is the integral of the previous lesson. A second application finishes it. In general $\int x^{n}e^{x}dx$ needs exactly $n$ applications, because each one differentiates $u$ once and nothing else reduces the power.
The signs alternate. The formula subtracts the new integral, so the second application's terms come in with the opposite sign to the first's. Written out, $$\int x^{2}e^{x}dx = x^{2}e^{x} - 2xe^{x} + 2e^{x} + C,$$ and a learner who loses one of those minus signs has done every piece of calculus correctly and produced a wrong answer. Laying the work out in two columns — derivatives on one side, integrals on the other, signs alternating from a plus — is the cheap protection against it.
Some integrals never terminate, and that is the point. In $\int e^{x}\sin x\,dx$ neither factor simplifies: differentiating a sine gives a cosine, and differentiating that gives a sine again. Two applications bring the original integral back:
$$I = e^{x}\sin x - e^{x}\cos x - I.$$
That is not a dead end. It is a linear equation in $I$, so $2I$ is the right-hand side and $I$ is half of it. The rule is worth stating plainly: when the integral you started with reappears, stop integrating and start doing algebra.
Another way: picture
Picture the power of $x$ as a flight of stairs. Each application of parts takes you down one step, and $x^{n}$ is $n$ steps from the ground. Now picture $e^{x}\sin x$ as a revolving door: every push moves you, two pushes bring you back to where you began, and the way out is not to push harder but to notice you are back.
Another way: steps
| The integrand | Applications | Why |
|---|---|---|
| $x e^{x}$ | 1 | the power drops to zero in one step |
| $x^{2}\sin x$ | 2 | two steps from $x^{2}$ to a constant |
| $x^{n}e^{x}$ | $n$ | one step per application, always |
| $e^{x}\sin x$ | never | nothing simplifies; it returns to itself |
The last row is the only one where more effort does not help, and it is the only one where the answer arrives by rearrangement.
A definite integral carries the limits through the $uv$ term as well, and in practice that term is frequently zero at both ends. On $[0, 1]$ with $u = x^{2}$ and $dv = (1-x)^{n}dx$, the $x^{2}$ kills it at the left end and the $(1-x)^{n+1}$ kills it at the right.
It is worth checking for that before doing any integration, because when it happens a repeated-parts calculation collapses to a single easy integral. The habit to build is: write $uv$, evaluate it at both limits, and only then decide how much work is left.
Losing the alternating sign. The second application's terms carry the opposite sign to the first's, because the formula subtracts the new integral every time. This is the error, far more than any other.
Stopping at the wrong moment. After one application of parts on $\int x^{2}e^{x}dx$ there is still an integral on the page. It is easier than the one you started with, and it is not the answer.
Treating a circular integral as a failure. $\int e^{x}\sin x\,dx$ coming back to itself is exactly what lets it be evaluated. A third application undoes the second and gets you nowhere; the equation does the rest.
$\int x^{2}e^{x}dx$: take $u = x^{2}$, $dv = e^{x}dx$, giving $x^{2}e^{x} - \int 2xe^{x}dx$.
One step down the stairs.
The remaining integral is the previous lesson's: $\int 2xe^{x}dx = 2xe^{x} - 2e^{x}$.
Second application.
So the answer is $x^{2}e^{x} - 2xe^{x} + 2e^{x} + C$ — note the signs running plus, minus, plus.
$I = \int e^{x}\sin x\,dx$: parts with $u = \sin x$ gives $e^{x}\sin x - \int e^{x}\cos x\,dx$.
Nothing got simpler.
Parts again on the new integral gives $e^{x}\cos x + \int e^{x}\sin x\,dx$, so $I = e^{x}\sin x - e^{x}\cos x - I$.
The original is back.
Therefore $2I = e^{x}(\sin x - \cos x)$ and $I = \tfrac{1}{2}e^{x}(\sin x - \cos x) + C$.
Take $u = x^{2}$ and $dv = \sin x\,dx$, so $v = -\cos x$ and the first application gives $-x^{2}\cos x + \int 2x\cos x\,dx$.
Watch the sign from integrating the sine.
Apply parts again to $\int 2x\cos x\,dx$: it gives $2x\sin x + 2\cos x$.
Second step down.
So the answer is $-x^{2}\cos x + 2x\sin x + 2\cos x + C$, and differentiating it back is the check worth doing.
Put the steps of evaluating $\int 2e^{x}\sin x\,dx$ into the order you do them.
Number the steps in order (write the number in the box):
Lay out $\int 4x^{2}e^{x}\,dx$ for repeated parts. Give the coefficient in each row of the differentiated column, and the sign that row carries.
| Coefficient | Sign | |
|---|---|---|
| First row | ||
| Second row | ||
| Third row | ||
| Fourth row |
Two applications of parts give $\int x^{2}e^{5x}\,dx = e^{5x}\left(\dfrac{x^{2}}{A} - \dfrac{2x}{B} + \dfrac{2}{C}\right) + K$. Give $A$, then $B$, then $C$.
A = p, B = q, C = w
Match each integral to how many applications of parts it takes to finish.
| One application | Two applications | Three applications | It never terminates; it returns to itself | |
|---|---|---|---|---|
| $\int 8x e^{x}\,dx$ | ||||
| $\int 8x^{2}e^{x}\,dx$ | ||||
| $\int 8x^{3}e^{x}\,dx$ | ||||
| $\int 8e^{x}\sin x\,dx$ |
Evaluate $\displaystyle\int_0^1 x^{2}(1 - x)^{2}\,dx$.
Answer:
Two applications of parts on $\int 2e^{x}\cos x\,dx$ have produced $I = 2e^{x}\cos x + 2e^{x}\sin x - I$, where $I$ is the original integral. What now?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of evaluating $\int 6e^{x}\sin x\,dx$ into the order you do them.
Number the steps in order (write the number in the box):
You can apply parts repeatedly and finish a circular integral. Without looking: how many applications does a power of $n$ need, what happens to the signs, and what do you do when the original integral reappears on the right?
10. Your turn: $\int x^{2}\sin x\,dx$, step 3