Integration by Parts

Main idea

Integration by parts is the product rule run backwards. It lets you move the derivative from one factor to the other, trading an integral you can't do, $\int u\,dv$ , for one you hopefully can, $\int v\,du$ .

\int {\htmlData{sym=0}{u}}\,{\htmlData{sym=1}{dv}} = {\htmlData{sym=4}{uv}} - \int {\htmlData{sym=3}{v}}\,{\htmlData{sym=2}{du}}

What the symbols mean

$u$
The part you differentiate
Pick the factor that gets simpler when differentiated, like x, x² or ln x.
$dv$
The part you integrate
Everything else, including dx. You must be able to integrate it.
$du$
Derivative of u
u′ dx. Usually simpler than u.
$v$
Antiderivative of dv
What you get by integrating dv. No + C needed here.
$uv$
Boundary term
What the product rule leaves over. For definite integrals, evaluate it at the limits.

Explanation

Integration by parts is simply a way to integrate a product of two things by shifting the hard work from one factor onto the other. The name means what it says: you split the integrand into two parts, one you'll differentiate and one you'll integrate. It's good for products like $xe^x$ , $x\sin x$ or $x^2\ln x$ , which substitution can't touch. All we're really saying with $\int u\,dv = uv - \int v\,du$ is: $\int u\,dv$ (the integral you're stuck on) equals $uv$ (a bonus term you get without integrating anything) minus $\int v\,du$ (a new integral you hope is easier). It looks like alphabet soup, but it's really one swap; bear with it.

Substitution was the chain rule run backwards. So what undoes the product rule? Start from it, $(uv)' = u'v + uv'$ , integrate both sides, and rearrange:

\int u\,v'\,dx = uv - \int u'\,v\,dx.

Writing $dv = v'\,dx$ and $du = u'\,dx$ gives the formula. Notice that nothing has actually been integrated yet. All you've done is move the derivative off $v$ and onto $u$ , at the cost of a boundary term $uv$ and a minus sign. That's only useful if the new integral is easier than the old one, so the whole game is choosing $u$ well.

Think about what makes a good $u$ . You're going to differentiate it, so you want something that gets simpler when differentiated: $x^2$ drops to $2x$ , and $\ln x$ turns into $\frac1x$ , which isn't even a log anymore. And you're going to integrate $dv$ , so it had better be something you can integrate without making it worse.

Choosing $u$ . A good rule of thumb is LIATE: pick $u$ from the earliest type in the list

Logarithms, like $\ln x$
Inverse trig, like $\arctan x$
Algebraic, like $x^2$ or $x$
Trig, like $\sin x$
Exponential, like $e^x$

and put everything else in $dv$ . The order isn't magic. Logs and inverse trig functions have simple derivatives but awkward antiderivatives, so you want to differentiate them. $e^x$ and $\sin x$ integrate without any fuss, so they're happy to sit in $dv$ .

Definite integrals. Same formula, with the boundary term evaluated at the limits:

\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du.

Repeat or solve. Sometimes one round isn't enough. For $\int x^2 e^x\,dx$ you apply parts twice, and each round knocks the power of $x$ down by one until it's gone. For $\int e^x \sin x\,dx$ something stranger happens: two rounds bring back the integral you started with. That's not a dead end. Call it $I$ , and you've got an equation like $I = (\text{stuff}) - I$ that you can solve for $I$ with plain algebra.

A common mistake. It's easy to lose the minus sign in front of $\int v\,du$ , since everything else in the formula is a plus. Another trap is picking $u = e^x$ in something like $\int xe^x\,dx$ because it's the "fancier" factor. Then $v = \frac{x^2}{2}$ , and the new integral $\int \frac{x^2}{2}e^x\,dx$ is worse than the one you started with. If the power of $x$ goes up, swap your choices.

There's an animation further down that shows the formula as a picture: a rectangle of area $uv$ split into the two areas that the formula trades between.

See it

The big rectangle has area uv, and the curve cuts it into two pieces: one is ∫u dv and the other is ∫v du. So if you know one piece, you get the other by subtracting it from uv. That subtraction is the whole formula.

Customers show up at a help desk at random, λ per minute on average, so the wait for the next one has density λe^(−λt). The average wait is the area under t·λe^(−λt), a power times an exponential, which is exactly the kind of integral parts was made for, and it comes out to 1/λ. As you slide λ from 0.5 up to 2, notice the area drops from 2 minutes to half a minute, while the curve's peak stays at the same height, about 0.37.

Worked examples

Example 1

Find $\displaystyle\int x e^x\,dx$ .

1
Algebraic comes before exponential in LIATE, so $u = x$ and $dv = e^x\,dx$ . Differentiating $x$ gets rid of it, which is what we want.

Example 2

Evaluate $\displaystyle\int_1^e \ln x\,dx$ .

1
It looks like there's only one factor, but you can always take $dv = dx$ . So let $u = \ln x$ ; its derivative is much nicer than the log itself.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $\displaystyle\int x\cos x\,dx$ .

Problem 2+5 XP

Find $\displaystyle\int x^2 e^{-x}\,dx$ .

Problem 3+5 XP

Evaluate $\displaystyle\int_0^{\pi/3} x\sin x\,dx$ .

Problem 4+5 XP

Find $\displaystyle\int \arcsin x\,dx$ .

Problem 5+5 XP

Find $\displaystyle\int e^{2x}\cos 3x\,dx$ .

Problem 6+5 XP

In a Fourier series, the first sine coefficient of a sawtooth wave $f(x) = x$ on $[-\pi, \pi]$ is $b_1 = \frac1\pi\int_{-\pi}^{\pi}x\sin x\,dx$ . Find it.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find $\displaystyle\int\sin(\ln x)\,dx$ .

Where it's used

In probability, the average waiting time of an exponential process is $\int_0^\infty t\,\lambda e^{-\lambda t}\,dt = 1/\lambda$ , found by parts. The same move turns the differential equations of structural engineering into the "weak form" that finite-element software solves, and it computes the Fourier coefficients that break a signal into frequencies.

Fourier SeriesCalc II · Electrical Engineering

Integrals Involving Trig Functions