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Lesson 85 — Integration by Parts

∫ u dv = uv − ∫ v du. Inverse of the product rule. LIATE heuristic for choosing u. Tabular method for polynomial × function.

Used in: Calculus II (Brazil) · Equiv. Math III Japanese · Equiv. Analysis LK German · AP Calculus BC (USA)

\int u\, dv = uv - \int v\, du

Integration by parts: inverse of the product rule. Choose u that simplifies upon differentiation and dv that has a known antiderivative v. LIATE heuristic: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — whichever appears first becomes u.

Choose your door

Rigorous notation, full derivation, hypotheses

Derivation, LIATE, and tabular method

Derivation of the formula

Definition· Integration by Parts

Let $u(x)$ and $v(x)$ be differentiable functions on $[a, b]$ , with $u'$ and $v'$ continuous. The product rule states:

$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$

Integrating both sides from $a$ to $b$ and applying the Fundamental Theorem of Calculus:

$[u(x)v(x)]_a^b = \int_a^b u'(x)v(x)\, dx + \int_a^b u(x)v'(x)\, dx.$

Rearranging:

$\int_a^b u(x)v'(x)\, dx = [u(x)v(x)]_a^b - \int_a^b v(x)u'(x)\, dx.$

In differential notation ( $dv = v'(x)\, dx$ , $du = u'(x)\, dx$ ):

\int u\, dv = uv - \int v\, du.

what this means · Formula for integration by parts — indefinite form.

"The formula for integration by parts comes from the product rule for differentiation: if $u$ and $v$ are both functions of $x$ , then $(uv)' = u'v + uv'$ . Integrating both sides and rearranging gives $\int u\, dv = uv - \int v\, du$ ." — Active Calculus §5.4

LIATE heuristic

"A useful heuristic for deciding which function to call $u$ in integration by parts is the acronym LIATE, which stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential functions." — OpenStax Calculus Vol. 2, §3.1

Tabular method (DI Method)

For integrals of the form $\int P(x) f(x)\, dx$ with $P$ a polynomial:

Tabular DI method for $\int x^2 e^x\, dx$ . Diagonal sums with alternating signs. Stop when D = 0.

Theorem· Trick: integral that returns to itself

For $I = \int e^x \cos x\, dx$ :

Apply parts with $u = \cos x$ , $dv = e^x dx$ : $I = e^x \cos x + \int e^x \sin x\, dx$ .

Apply parts again with $u = \sin x$ : $I = e^x \cos x + e^x \sin x - I$ .

Therefore $2I = e^x(\cos x + \sin x)$ , and:

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

what this means · Result of the trick when the original integral reappears after two applications.

Condition: the trick works when the second application reproduces the original integral with the same sign (not inverted, which would give a null identity).

Worked examples

Example— 3· Polynomial x² with DI table (intermediate)

Problem. Compute $\int x^2 e^x\, dx$ using the tabular method.

Strategy. Polynomial of degree 2 × exponential: DI method (differentiate $x^2$ to zero, integrate $e^x$ repeatedly).

Solution.

D	I	Sign
$x^2$	$e^x$	+
$2x$	$e^x$	−
$2$	$e^x$	+
$0$	—	—

Summing diagonal products with alternating signs:

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

Verification. Differentiating $e^x(x^2 - 2x + 2)$ : $e^x(x^2 - 2x + 2) + e^x(2x - 2) = e^x x^2$ . Correct.

Source. APEX Calculus §6.2, Example 6.2.4 — CC-BY-NC.

Example— 4· Trick: integral that returns (e^x cos x)

Problem. Compute $I = \int e^x \cos x\, dx$ .

Strategy. Two applications of parts — the integral will reappear. Isolate $I$ .

Solution.

1st application: $u = \cos x$ , $dv = e^x\, dx$ , $du = -\sin x\, dx$ , $v = e^x$ .

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

2nd application: $u = \sin x$ , $dv = e^x\, dx$ , $du = \cos x\, dx$ , $v = e^x$ .

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

Therefore $2I = e^x(\cos x + \sin x)$ :

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

Verification. Differentiating: $\frac{e^x(\sin x + \cos x) + e^x(\cos x - \sin x)}{2} = \frac{2e^x \cos x}{2} = e^x \cos x$ . Correct.

Source. OpenStax Calculus Vol. 2, §3.1, Example 3.3 — CC-BY-NC-SA.

Example— 5· Definite integral with parts (challenge)

Problem. Compute $\int_0^\pi x \sin x\, dx$ .

Strategy. LIATE: $u = x$ , $dv = \sin x\, dx$ . Apply parts and evaluate at limits.

Solution.

$du = dx$ , $v = -\cos x$ .

$\int_0^\pi x \sin x\, dx = [-x \cos x]_0^\pi + \int_0^\pi \cos x\, dx.$

$[-x \cos x]_0^\pi = -\pi\cos\pi - 0 = -\pi(-1) = \pi$ .

$\int_0^\pi \cos x\, dx = [\sin x]_0^\pi = 0 - 0 = 0$ .

Therefore $\int_0^\pi x \sin x\, dx = \pi + 0 = \pi$ .

Verification. The result $\pi$ is classical — confirmed numerically ( $\approx 3.1416$ ). The asymmetry of the sign of $x\sin x$ on $[0, \pi]$ (always positive) guarantees a positive result.

Source. Active Calculus §5.4, Exercise 5.4.3 — CC-BY-NC-SA.

Exercise list

45 exercises · 11 with worked solution (25%)

Application 30Modeling 8Challenge 5Proof 2

Ex. 85.1Application
Compute $\int x e^x\, dx$ .
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Choose $u = x$ , $dv = e^x dx$ . Then $du = dx$ , $v = e^x$ . Applying the formula: $xe^x - \int e^x dx = xe^x - e^x + C = e^x(x-1) + C$ .
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1. Identify the product $x \cdot e^x$ . LIATE: $x$ is Algebraic, $e^x$ is Exponential. Thus $u = x$ .
2. Then $dv = e^x dx$ . Compute: $du = dx$ , $v = e^x$ .
3. Apply $\int u\, dv = uv - \int v\, du$ : $xe^x - \int e^x dx$ .
4. Solve the remaining integral: $\int e^x dx = e^x$ .
5. Result: $xe^x - e^x + C = e^x(x-1) + C$ . Tip: always verify by differentiating the result.
Ex. 85.2Application
Compute $\int x \sin x\, dx$ .
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$u = x$ , $dv = \sin x\, dx$ , $du = dx$ , $v = -\cos x$ . Result: $-x\cos x + \int \cos x\, dx = -x\cos x + \sin x + C$ .
Ex. 85.3Application
Compute $\int x \cos x\, dx$ .
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$u = x$ , $dv = \cos x\, dx$ , $v = \sin x$ . Result: $x\sin x - \int \sin x\, dx = x\sin x + \cos x + C$ .
Ex. 85.4Application
Compute $\int \ln x\, dx$ .
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LIATE: $u = \ln x$ , $dv = dx$ . Then $du = dx/x$ , $v = x$ . Result: $x\ln x - \int 1\, dx = x\ln x - x + C$ .
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1. $\ln x$ has no accompanying factor. Use the trick: $dv = dx$ (trivially integrates to $v = x$ ).
2. $u = \ln x$ , $du = dx/x$ .
3. Apply the formula: $x\ln x - \int x \cdot (1/x)\, dx = x\ln x - \int 1\, dx$ .
4. Result: $x\ln x - x + C$ . Tip: whenever the integrand is just a logarithm or inverse trig, use dv = dx.
Ex. 85.5Application
Compute $\int x^2 e^x\, dx$ . Apply parts twice or use the tabular method.
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Apply parts twice or use DI table. $u = x^2$ , $dv = e^x dx$ . After two cycles: $x^2 e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C$ .
Ex. 85.6Application
Compute $\int x^2 \sin x\, dx$ .
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DI table with $D = x^2$ , $I = \sin x$ : $-x^2\cos x + 2x\sin x + 2\cos x + C$ .
Ex. 85.7ApplicationAnswer key
Compute $\int x^2 \cos x\, dx$ .
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DI table: $x^2 \sin x + 2x\cos x - 2\sin x + C$ .
Ex. 85.8ApplicationAnswer key
Compute $\int x \ln x\, dx$ .
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$u = \ln x$ , $dv = x\, dx$ , $v = x^2/2$ . Result: $\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$ .
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1. LIATE: $\ln x$ is Log, $x$ is Algebraic. Thus $u = \ln x$ .
2. $dv = x\, dx$ , hence $v = x^2/2$ ; $du = dx/x$ .
3. Apply: $\frac{x^2}{2}\ln x - \int \frac{x^2}{2} \cdot \frac{1}{x}\, dx = \frac{x^2}{2}\ln x - \frac{1}{2}\int x\, dx$ .
4. Result: $\frac{x^2}{2}\ln x - \frac{x^2}{4} + C$ . Insight: general pattern is ∫ x^n ln x dx = x^(n+1) ln x/(n+1) − x^(n+1)/(n+1)².
Ex. 85.9Application
Compute $\int x^2 \ln x\, dx$ .
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$u = \ln x$ , $dv = x^2\, dx$ . Result: $\frac{x^3}{3}\ln x - \frac{x^3}{9} + C$ .
Ex. 85.10ApplicationAnswer key
Compute $\int \arctan x\, dx$ .
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$u = \arctan x$ , $dv = dx$ , $du = dx/(1+x^2)$ , $v = x$ . Result: $x\arctan x - \frac{1}{2}\ln(1+x^2) + C$ .
Ex. 85.11Application
Compute $\int \arcsin x\, dx$ .
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$u = \arcsin x$ , $dv = dx$ , $du = dx/\sqrt{1-x^2}$ , $v = x$ . Result: $x\arcsin x + \sqrt{1-x^2} + C$ .
Ex. 85.12Application
Compute $\int x e^{2x}\, dx$ .
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$u = x$ , $dv = e^{2x}\, dx$ , $v = e^{2x}/2$ . Result: $\frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C$ .
Ex. 85.13Application
Compute $\int x \cdot 2^x\, dx$ .
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$u = x$ , $dv = 2^x\, dx$ , $v = 2^x/\ln 2$ . Result: $\frac{x \cdot 2^x}{\ln 2} - \frac{2^x}{(\ln 2)^2} + C$ .
Ex. 85.14Application
Compute $\int x e^{-x}\, dx$ .
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$u = x$ , $dv = e^{-x}\, dx$ , $v = -e^{-x}$ . Result: $-xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C$ .
Ex. 85.15ApplicationAnswer key
Compute $\int (\ln x)^2\, dx$ . Apply parts twice.
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Apply parts with $u = (\ln x)^2$ , $dv = dx$ . After two cycles: $x(\ln x)^2 - 2x\ln x + 2x + C$ .
Ex. 85.16Application
Compute $\int e^x \cos x\, dx$ .
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Two applications with trick. Result: $\frac{e^x(\sin x + \cos x)}{2} + C$ .
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1. Define $I = \int e^x \cos x\, dx$ . Apply parts: $u = \cos x$ , $dv = e^x\, dx$ .
2. Result of 1st application: $I = e^x \cos x + \int e^x \sin x\, dx$ .
3. Apply parts again to the new integral: $u = \sin x$ , $dv = e^x\, dx$ . Obtain $e^x \sin x - \int e^x \cos x\, dx = e^x \sin x - I$ .
4. Substitute: $I = e^x \cos x + e^x \sin x - I$ , hence $2I = e^x(\cos x + \sin x)$ .
5. Result: $I = \frac{e^x(\sin x + \cos x)}{2} + C$ . Tip: when the integral returns, isolate I — it is a grave error to cancel both sides as zero.
Ex. 85.17Application
Compute $\int e^x \sin x\, dx$ .
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Analogous trick to 85.16: $\int e^x \sin x\, dx = \frac{e^x(\sin x - \cos x)}{2} + C$ .
Ex. 85.18Application
Compute $\int e^{2x} \cos(3x)\, dx$ .
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Trick with $a = 2$ , $b = 3$ . Result: $\frac{e^{2x}(2\cos 3x + 3\sin 3x)}{13} + C$ .
Ex. 85.19Application
Compute $\int e^{-x} \sin(2x)\, dx$ .
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Result: $\frac{e^{-x}(-\sin 2x - 2\cos 2x)}{5} + C$ .
Ex. 85.20Application
Compute $\int \cos x \ln(\sin x)\, dx$ .
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Sub $t = \sin x$ first, then parts in $\int \ln t\, dt$ . Result: $\sin x \ln(\sin x) - \sin x + C$ .
Ex. 85.21Application
Compute $\int \sec^3 x\, dx$ .
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Parts with $u = \sec x$ , $dv = \sec^2 x\, dx$ plus identity $\sec^2 = 1 + \tan^2$ . Result: $\frac{\sec x \tan x}{2} + \frac{1}{2}\ln|\sec x + \tan x| + C$ .
Ex. 85.22ApplicationAnswer key
Compute $\int \csc^3 x\, dx$ .
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Analogous to sec³: $\int \csc^3 x\, dx = -\frac{\csc x \cot x}{2} - \frac{1}{2}\ln|\csc x + \cot x| + C$ .
Ex. 85.23ApplicationAnswer key
Compute $\int_0^1 x e^x\, dx$ .
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From indefinite $e^x(x-1)$ , evaluate on $[0,1]$ : $[e^x(x-1)]_0^1 = e^1 \cdot 0 - e^0(-1) = 0 + 1 = 1$ .
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1. Compute the antiderivative: $\int x e^x\, dx = e^x(x-1) + C$ (see 85.1).
2. Evaluate at limits: $[e^x(x-1)]_0^1$ .
3. At $x=1$ : $e^1(1-1) = 0$ . At $x=0$ : $e^0(0-1) = -1$ .
4. Result: $0 - (-1) = 1$ . Insight: area under xe^x on [0,1] is exactly 1, an elegant result.
Ex. 85.24Application
Compute $\int_0^\pi x \sin x\, dx$ .
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From indefinite $-x\cos x + \sin x$ , evaluate on $[0,\pi]$ : $[-x\cos x + \sin x]_0^\pi = \pi + 0 - 0 = \pi$ .
Ex. 85.25Application
Compute $\int_1^e \ln x\, dx$ .
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From indefinite $x\ln x - x$ , evaluate on $[1,e]$ : $[x\ln x - x]_1^e = (e - e) - (0 - 1) = 1$ .
Ex. 85.26Application
Compute $\int_0^{\pi/2} x \cos x\, dx$ .
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From indefinite $x\sin x + \cos x$ , evaluate on $[0, \pi/2]$ : $[x\sin x + \cos x]_0^{\pi/2} = \pi/2 + 0 - 1 = \pi/2 - 1$ .
Ex. 85.27Application
Compute $\int_0^1 \arctan x\, dx$ .
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From indefinite $x\arctan x - \frac{1}{2}\ln(1+x^2)$ , evaluate on $[0,1]$ : $\frac{\pi}{4} - \frac{\ln 2}{2}$ .
Ex. 85.28Application
Compute $\int_1^2 x^2 \ln x\, dx$ .
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From indefinite $\frac{x^3}{3}\ln x - \frac{x^3}{9}$ , evaluate on $[1,2]$ : $\frac{8\ln 2}{3} - \frac{8}{9} + \frac{1}{9} = \frac{8\ln 2}{3} - \frac{7}{9}$ .
Ex. 85.29Application
Compute $\int_0^1 x e^{-x}\, dx$ .
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From indefinite $-e^{-x}(x+1)$ , evaluate on $[0,1]$ : $-2e^{-1} - (-1) = 1 - 2/e$ .
Ex. 85.30Application
Compute $\int_0^{\pi/2} e^x \sin x\, dx$ .
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From indefinite $\frac{e^x(\sin x - \cos x)}{2}$ , evaluate on $[0, \pi/2]$ : $\frac{e^{\pi/2} + 1}{2}$ .
Ex. 85.31ModelingAnswer key
Work done by force $F(x) = x e^{-x}$ N along $[0, 1]$ m. Compute $W = \int_0^1 F(x)\, dx$ .
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Work $W = \int_0^1 x e^{-x}\, dx = 1 - 2e^{-1} \approx 0.26$ J. (See 85.29.)
Ex. 85.32ModelingAnswer key
Electric charge accumulated with current $i(t) = t\cos(\omega t)$ . Compute $Q = \int_0^{2\pi/\omega} i(t)\, dt$ .
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$\int_0^{2\pi/\omega} t\cos(\omega t)\, dt$ . By parts: $u = t$ , $dv = \cos(\omega t)\, dt$ . Evaluating at limits: result is $0$ (cosine symmetry over full period).
Ex. 85.33Modeling
Gamma function: compute $\Gamma(2) = \int_0^\infty t e^{-t}\, dt$ and show that the result is $1$ .
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$\Gamma(2) = \int_0^\infty t e^{-t}\, dt$ . By parts: $u = t$ , $dv = e^{-t}\, dt$ . Result: $[-te^{-t}]_0^\infty + \int_0^\infty e^{-t}\, dt = 0 + 1 = 1$ .
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Ex. 85.34ModelingAnswer key
Expectation of exponential variable: compute $E[X] = \int_0^\infty x \lambda e^{-\lambda x}\, dx$ and show that the result is $1/\lambda$ .
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$E[X] = \int_0^\infty x\lambda e^{-\lambda x}\, dx$ . By parts: $\lambda\left([-xe^{-\lambda x}/1]_0^\infty + \frac{1}{\lambda}\int_0^\infty e^{-\lambda x}\, dx\right) = \lambda \cdot \frac{1}{\lambda^2} = \frac{1}{\lambda}$ .
Ex. 85.35Modeling
Variance of exponential: compute $E[X^2] = \int_0^\infty x^2 \lambda e^{-\lambda x}\, dx$ and determine $\text{Var}(X) = E[X^2] - (E[X])^2$ .
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$E[X^2] = \int_0^\infty x^2 \lambda e^{-\lambda x}\, dx$ . DI table: result $2/\lambda^2$ . Hence variance $= E[X^2] - (E[X])^2 = 2/\lambda^2 - 1/\lambda^2 = 1/\lambda^2$ .
Ex. 85.36Modeling
Fourier coefficient $a_n = (1/\pi)\int_{-\pi}^\pi x\cos(nx)\, dx$ . Determine $a_n$ for all integers $n \geq 1$ .
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By parts: $a_n = \frac{1}{\pi}\int_{-\pi}^\pi x\cos(nx)\, dx$ . Since $x\cos(nx)$ is product of odd function ( $x$ ) and even ( $\cos(nx)$ ), it is odd. Integral of odd function over symmetric interval is $0$ . Hence $a_n = 0$ for all $n$ .
Ex. 85.37Modeling
Laplace transform of $t$ : show that $\mathcal{L}\{t\}(s) = \int_0^\infty t e^{-st}\, dt = 1/s^2$ for $s > 0$ .
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$\mathcal{L}\{t\} = \int_0^\infty t e^{-st}\, dt$ . By parts: $u = t$ , $dv = e^{-st}\, dt$ . Result: $[-t e^{-st}/s]_0^\infty + \frac{1}{s}\int_0^\infty e^{-st}\, dt = 0 + \frac{1}{s^2} = \frac{1}{s^2}$ .
Ex. 85.38Modeling
Present value of growing income $C(t) = t$ (in thousands of reais/year) with continuous discount rate $r$ over $[0, T]$ . Compute $VP = \int_0^T t e^{-rt}\, dt$ .
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PV = $\int_0^T t e^{-rt}\, dt$ . By parts: $u = t$ , result: $\frac{1}{r^2}[1 - e^{-rT}(1 + rT)]$ .
Ex. 85.39Challenge
Compute $\int x^3 e^{-x^2}\, dx$ . Hint: substitute $u = x^2$ first, then apply parts.
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Sub $u = x^2$ : $\int x^3 e^{-x^2}\, dx = \frac{1}{2}\int u e^{-u}\, du$ . By parts: $-\frac{1}{2}e^{-x^2}(x^2 + 1) + C$ .
Ex. 85.40ChallengeAnswer key
Compute $\int e^{\sqrt{x}}\, dx$ . Hint: substitute $u = \sqrt{x}$ , then apply parts.
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Sub $u = \sqrt{x}$ , $x = u^2$ , $dx = 2u\, du$ : integral becomes $2\int u e^u\, du$ . By parts: $2(u-1)e^u + C = 2(\sqrt{x} - 1)e^{\sqrt{x}} + C$ .
Ex. 85.41Challenge
Show by induction that $\int_0^\infty x^n e^{-x}\, dx = n!$ for all non-negative integer $n$ , using integration by parts in the inductive step.
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By parts: $\int_0^\infty x^n e^{-x}\, dx = n \int_0^\infty x^{n-1} e^{-x}\, dx$ . By induction: $= n(n-1)\cdots 1 \int_0^\infty e^{-x}\, dx = n!$ .
Ex. 85.42Challenge
Let $I_n = \int x^n e^x\, dx$ . Show that $I_n = x^n e^x - n I_{n-1}$ and use the formula to compute $I_3$ .
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By parts with $u = x^n$ , $dv = e^x\, dx$ : $I_n = x^n e^x - n\int x^{n-1}e^x\, dx = x^n e^x - nI_{n-1}$ .
Ex. 85.43ChallengeAnswer key
Derive the reduction formula $\int \sin^n x\, dx = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}\int \sin^{n-2}x\, dx$ and apply it to compute $\int \sin^4 x\, dx$ .
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By parts with $u = \sin^{n-1}x$ , $dv = \sin x\, dx$ . Result: $\int \sin^n x\, dx = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}\int \sin^{n-2}x\, dx$ . Use for $n = 4$ : $-\frac{\sin^3 x\cos x}{4} + \frac{3}{4}\int \sin^2 x\, dx$ .
Ex. 85.44Proof
Proof. Prove the formula $\int u\, dv = uv - \int v\, du$ from the product rule for derivatives and the Fundamental Theorem of Calculus.
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By the product rule: $(uv)' = u'v + uv'$ . Integrate both sides on $[a,b]$ and apply FTC: $[uv]_a^b = \int_a^b u'v\, dx + \int_a^b uv'\, dx$ . Rearranging: $\int_a^b uv'\, dx = [uv]_a^b - \int_a^b u'v\, dx$ , which in differential notation is $\int u\, dv = uv - \int v\, du$ .
Ex. 85.45Proof
Informal Proof. Justify why LIATE is an effective heuristic: analyze how each type of function behaves upon differentiation and explain why Log and Inv-trig are the top candidates for $u$ .
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See the referenced source for the detailed solution.

Sources

Active Calculus 2.0 — Boelkins · 2024 · CC-BY-NC-SA · §5.4. Primary source.
Calculus Volume 2 (OpenStax) — OpenStax · 2016 · CC-BY-NC-SA · §3.1.
APEX Calculus v5 — Hartman et al. · 2024 · CC-BY-NC · §6.2–6.3.