Difficulty: Intermediate | Prerequisites: Basic integration, product rule for differentiation, natural logarithm properties
Big picture: Integration by parts is the integration counterpart of the product rule for derivatives. It is the go-to technique whenever your integrand is a product of two different types of function (a polynomial times a logarithm, a polynomial times a trig function, an exponential times a trig function). This homework also revisits u-substitution and the Fundamental Theorem of Calculus with the chain rule, tying together the core toolkit you will carry through the rest of the course.
Integration by parts turns a hard integral of a product into a simpler one by choosing one factor to differentiate and the other to integrate. The classic formula is: integral of u dv = uv minus integral of v du. Most of the problems here involve pairing a polynomial or logarithm with a trig function or exponential, and the trick is knowing which factor to call u (usually the one that simplifies when you differentiate it).
Integration by parts (IBP)
The rule that integral of u dv = uv - integral of v du. In simple terms, you trade one integral for another by differentiating one piece and integrating the other.
LIATE rule
A mnemonic for choosing u in integration by parts. Pick u from the list in this order of preference: Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. The idea is that functions earlier in the list tend to simplify when differentiated.
Tabular method (repeated IBP)
A shortcut for integration by parts when one factor is a polynomial. You build a table of successive derivatives of u and successive antiderivatives of dv, alternating signs. Useful when IBP must be applied multiple times.
Cycling (IBP with exponentials and trig)
When integrating a product of an exponential and a trig function, two rounds of IBP produce the original integral again. You then solve algebraically for the integral.
Fundamental Theorem of Calculus with the chain rule
If F(x) = d/dx of integral from 0 to g(x) of f(t) dt, then F(x) = f(g(x)) times g'(x). In simple terms, you evaluate the integrand at the upper limit and multiply by the derivative of that limit.
u-substitution
The integration counterpart of the chain rule. If the integrand contains a composite function and its derivative, substituting u for the inner function simplifies the integral.
Logarithms cannot be integrated directly in a simple way, so when ln(x) appears in a product, it is almost always chosen as u (the part you differentiate), because d/dx of ln(x) = 1/x, which simplifies nicely.
Worked example (HW Q11): integral from 1 to 2 of 4 ln(3x) dx
Let u = ln(3x), dv = 4 dx. Then du = 1/x dx, v = 4x.
IBP gives: [4x ln(3x)] from 1 to 2 - integral from 1 to 2 of 4 dx = (8 ln 6 - 4 ln 3) - 4.
Simplify: 8 ln 6 - 4 ln 3 - 4 = 4(2 ln 6 - ln 3 - 1) = 4(ln 36 - ln 3 - 1) = 4(ln 12 - 1).
Answer: I = 4(ln 12 - 1).
Worked example (HW Q12): integral from 1 to e of 4x ln(x) dx
Let u = ln(x), dv = 4x dx. Then du = 1/x dx, v = 2x^2.
IBP gives: [2x^2 ln(x)] from 1 to e - integral from 1 to e of 2x dx = (2e^2 - 0) - [x^2] from 1 to e = 2e^2 - (e^2 - 1) = e^2 + 1.
Answer: I = e^2 + 1.
Worked example (HW Q13): integral from 1 to e of 4x^4 ln(x) dx
Let u = ln(x), dv = 4x^4 dx. Then du = 1/x dx, v = 4x^5/5.
IBP gives: [4x^5 ln(x)/5] from 1 to e - integral from 1 to e of (4x^4/5) dx = (4e^5/5 - 0) - [4x^5/25] from 1 to e.
Evaluating: 4e^5/5 - (4e^5/25 - 4/25) = 4e^5/5 - 4e^5/25 + 4/25 = (20e^5 - 4e^5 + 4)/25 = (16e^5 + 4)/25 = 4(4e^5 + 1)/25.
Answer: I = (4/25)(4e^5 + 1).
Worked example (HW Q14): integral from e to 3 of ln(x)/x^2 dx
Let u = ln(x), dv = x^(-2) dx. Then du = 1/x dx, v = -1/x.
IBP gives: [-ln(x)/x] from e to 3 + integral from e to 3 of 1/x^2 dx = (-ln 3/3 + 1/e) + [-1/x] from e to 3.
Evaluating: (-ln 3/3 + 1/e) + (-1/3 + 1/e) = 2/e - ln 3/3 - 1/3 = 2/e - (1/3)(ln 3 + 1).
Answer: I = 2/e - (1/3)(ln 3 + 1).
Polynomial times trig: repeated IBP (tabular method)
When integrating x^n times cos(ax) or sin(ax), you apply IBP n times. The polynomial eventually differentiates to zero. The tabular method streamlines this.
Worked example (HW Q15): integral of (x^2 + 2) cos(2x) dx
Using the tabular method with u-column = x^2 + 2 and dv-column = cos(2x):
Derivatives of u: x^2 + 2, then 2x, then 2, then 0.
Antiderivatives of cos(2x): sin(2x)/2, then -cos(2x)/4, then -sin(2x)/8.
Alternating signs: +, -, +.
Result: (x^2 + 2)(sin(2x)/2) - (2x)(-cos(2x)/4) + (2)(-sin(2x)/8) + C.
Simplifying: (x^2 + 2) sin(2x)/2 + x cos(2x)/2 - sin(2x)/4 + C = (1/4)[2x cos(2x) + (2x^2 + 3) sin(2x)] + C.
Answer: I = (1/4)[2x cos(2x) + (2x^2 + 3) sin(2x)] + C.
Exponential times substitution (HW Q16): integral from 0 to 1 of e^(8 sqrt(x)) dx
First substitute: let t = sqrt(x), so x = t^2, dx = 2t dt. When x = 0, t = 0; when x = 1, t = 1.
Integral becomes: integral from 0 to 1 of e^(8t) times 2t dt.
Now apply IBP with u = 2t, dv = e^(8t) dt. Then du = 2 dt, v = e^(8t)/8.
IBP gives: [2t e^(8t)/8] from 0 to 1 - integral from 0 to 1 of (2/8) e^(8t) dt = e^8/4 - [e^(8t)/32] from 0 to 1 = e^8/4 - (e^8 - 1)/32.
Simplifying: (8e^8 - e^8 + 1)/32 = (7e^8 + 1)/32.
Answer: I = (1/32)(7e^8 + 1).
Exponential times trig: cycling (HW Q17): integral of e^(-x) cos(2x) dx
Apply IBP twice. Let u = cos(2x), dv = e^(-x) dx.
First IBP: -e^(-x) cos(2x) - integral of 2 e^(-x) sin(2x) dx.
Second IBP on the remaining integral (u = sin(2x), dv = e^(-x) dx): this yields -e^(-x) sin(2x) + integral of (-2) e^(-x) cos(2x) dx.
Putting it together, the original integral I appears again on the right side:
I = -e^(-x) cos(2x) - 2[-e^(-x) sin(2x) - 2I] = -e^(-x) cos(2x) + 2e^(-x) sin(2x) + 4I... wait, let me redo this carefully.
Let I = integral of e^(-x) cos(2x) dx.
First IBP: u = e^(-x), dv = cos(2x) dx. Then du = -e^(-x) dx, v = sin(2x)/2.
I = e^(-x) sin(2x)/2 + (1/2) integral of e^(-x) sin(2x) dx.
Second IBP on the new integral: u = e^(-x), dv = sin(2x) dx. Then du = -e^(-x) dx, v = -cos(2x)/2.
Integral of e^(-x) sin(2x) dx = -e^(-x) cos(2x)/2 - (1/2) integral of e^(-x) cos(2x) dx = -e^(-x) cos(2x)/2 - I/2.
Substituting back: I = e^(-x) sin(2x)/2 + (1/2)[-e^(-x) cos(2x)/2 - I/2] = e^(-x) sin(2x)/2 - e^(-x) cos(2x)/4 - I/4.
Solving: I + I/4 = e^(-x)(2 sin(2x) - cos(2x))/4. So 5I/4 = e^(-x)(2 sin(2x) - cos(2x))/4.
I = (1/5) e^(-x)(2 sin(2x) - cos(2x)) + C.
Answer: I = (1/5) e^(-x)(2 sin 2x - cos 2x) + C.
IBP with sec^2 (HW Q18): integral from 0 to pi/4 of (2x - 1) sec^2(x) dx
Let u = 2x - 1, dv = sec^2(x) dx. Then du = 2 dx, v = tan(x).
IBP gives: [(2x - 1) tan(x)] from 0 to pi/4 - integral from 0 to pi/4 of 2 tan(x) dx.
Evaluating the boundary term: (pi/2 - 1)(1) - 0 = pi/2 - 1.
The remaining integral: 2 integral of tan(x) dx = 2[-ln|cos(x)|] from 0 to pi/4 = 2[-ln(sqrt(2)/2) + 0] = 2 ln(sqrt(2)) ... let me be precise.
-2[ln|cos(x)|] from 0 to pi/4 = -2[ln(cos(pi/4)) - ln(cos(0))] = -2[ln(1/sqrt(2)) - 0] = -2(-ln(sqrt(2))) = 2 ln(sqrt(2)) = ln 2.
So I = pi/2 - 1 - ln 2.
Answer: I = pi/2 - 1 - ln 2.
IBP with abstract f''(x) (HW Q19)
Evaluate integral from 0 to 1 of x f''(x) dx, given f(0) = 5, f(1) = 3, f'(0) = 2, f'(1) = 4.
Let u = x, dv = f''(x) dx. Then du = dx, v = f'(x).
IBP gives: [x f'(x)] from 0 to 1 - integral from 0 to 1 of f'(x) dx = (1 times 4 - 0) - [f(x)] from 0 to 1 = 4 - (3 - 5) = 4 + 2 = 6.
Answer: I = 6.
IBP with x^2 f''(x) (HW Q20)
Evaluate integral from 1 to 2 of x^2 f''(x) dx, given f(1) = 9, f(2) = 3, f'(1) = 4, f'(2) = 1.
Let u = x^2, dv = f''(x) dx. Then du = 2x dx, v = f'(x).
First IBP: [x^2 f'(x)] from 1 to 2 - integral from 1 to 2 of 2x f'(x) dx = (4 times 1 - 1 times 4) - 2 integral from 1 to 2 of x f'(x) dx = 0 - 2 integral from 1 to 2 of x f'(x) dx.
Apply IBP again to integral of x f'(x) dx: u = x, dv = f'(x) dx, du = dx, v = f(x).
This gives: [x f(x)] from 1 to 2 - integral from 1 to 2 of f(x) dx = (6 - 9) - integral from 1 to 2 of f(x) dx = -3 - integral from 1 to 2 of f(x) dx.
So the original integral = -2(-3 - integral from 1 to 2 of f(x) dx) = 6 + 2 integral from 1 to 2 of f(x) dx.
Answer: I = 6 + 2 integral from 1 to 2 of f(x) dx.
FTC with chain rule (HW Q21)
F(x) = d/dx of integral from 0 to x^4 of 2t^5 dt. Find F(1).
By the FTC with chain rule: F(x) = 2(x^4)^5 times d/dx(x^4) = 2x^20 times 4x^3 = 8x^23.
F(1) = 8.
Answer: F(1) = 8.
u-substitution (HW Q22): integral of 8x(x^2/3 - 3)^3 dx
Let u = x^2/3 - 3. Then du = (2x/3) dx, so x dx = (3/2) du.
Substituting: 8 times (3/2) integral of u^3 du = 12 times u^4/4 = 3u^4 + C = 3(x^2/3 - 3)^4 + C.
Answer: I = 3(x^2/3 - 3)^4 + C.
\text{Integration by parts:} \quad \int u\, dv = uv - \int v\, du\text{FTC with chain rule:} \quad \frac{d}{dx}\int_0^{g(x)} f(t)\, dt = f\bigl(g(x)\bigr) \cdot g'(x)\text{u-substitution:} \quad \int f\bigl(g(x)\bigr) g'(x)\, dx = \int f(u)\, du \quad \text{where } u = g(x)Tabular method layout
Draw three columns: derivatives of u (going down), antiderivatives of dv (going down), and alternating signs (+, -, +, -, ...). Multiply diagonally and sum. Stop when the derivative column reaches zero.
Students often choose u and dv backwards. If you pick u = cos(x) and dv = x^2 dx when integrating x^2 cos(x), the integral gets harder, not simpler. Use LIATE: the polynomial is u.
When IBP cycles (exponential times trig), students sometimes apply IBP a third time instead of recognising that the original integral has reappeared. After two rounds, solve the resulting equation algebraically.
Students forget that IBP on a definite integral requires evaluating the boundary term [uv] at both limits, not just the integral part. Skipping this loses marks.
In Q19/Q20-style problems, students try to find f(x) explicitly. You do not need f(x). The given values of f, f', and f'' at the endpoints are enough to evaluate everything after IBP.
Integration by parts is tested on virtually every integral calculus exam. Expect at least two IBP problems, often one with logarithms and one with trig or exponentials.
The cycling trick (exponential times trig) is a classic exam question because it tests whether you can recognise a pattern and solve algebraically rather than continuing to integrate forever.
Problems involving abstract functions (where you are given values of f, f', f'' at specific points) test conceptual understanding of IBP, not just mechanical computation. These appear frequently on UT Austin exams.
The FTC with chain rule (Q21) is a short problem that examiners love because it tests whether you remember to multiply by the derivative of the upper limit.
Fill in the blank: In the LIATE rule, L stands for _____ and it should be your first choice for u. (Logarithmic)
True or false: When integrating e^x sin(x), two rounds of IBP will bring back the original integral, which you then solve for algebraically. _____ (True.)
Fill in the blank: d/dx of integral from 0 to x^3 of f(t) dt = _____ times _____. (f(x^3) times 3x^2)
True or false: When applying IBP to a definite integral, the [uv] boundary term is always zero. _____ (False. It is evaluated at both limits and is usually non-zero.)
Fill in the blank: In the tabular method, you stop taking derivatives when the u column reaches _____. (Zero)
Q: Evaluate integral of x e^(3x) dx.
A: Let u = x, dv = e^(3x) dx. Then du = dx, v = e^(3x)/3. IBP gives: x e^(3x)/3 - integral of e^(3x)/3 dx = x e^(3x)/3 - e^(3x)/9 + C = (e^(3x)/9)(3x - 1) + C.
Q: Evaluate integral from 1 to e of ln(x) dx.
A: Let u = ln(x), dv = dx. Then du = 1/x dx, v = x. IBP gives: [x ln(x)] from 1 to e - integral from 1 to e of dx = (e - 0) - (e - 1) = 1.
Q: If F(x) = d/dx of integral from 0 to x^2 of sin(t^3) dt, what is F(x)?
A: By the FTC with chain rule: F(x) = sin((x^2)^3) times 2x = 2x sin(x^6).
Q: Evaluate integral of e^(2x) sin(x) dx using the cycling method.
A: Apply IBP twice. After two rounds, the original integral reappears. Solving: I = e^(2x)(2 sin x - cos x)/5 + C.
Q: Use IBP to express integral from 0 to 1 of x^2 f'(x) dx in terms of f, given f(0) = 1, f(1) = 5.
A: Let u = x^2, dv = f'(x) dx. Then du = 2x dx, v = f(x). IBP gives: [x^2 f(x)] from 0 to 1 - 2 integral from 0 to 1 of x f(x) dx = 5 - 2 integral from 0 to 1 of x f(x) dx.
Integration by parts is essential for deriving reduction formulas (e.g. for integral of sin^n(x) dx), which you will meet in later chapters. It also underpins the Laplace transform in differential equations, where integrals of e^(-st) f(t) rely heavily on IBP. The FTC with chain rule connects to Leibniz's rule for differentiating under the integral sign, which appears in advanced calculus and physics.
Related Terms / Search Tags
integration by parts, IBP, LIATE rule, tabular integration, repeated integration by parts, cycling method, exponential times trig integral, reduction formula, Fundamental Theorem of Calculus chain rule, Leibniz integral rule, u-substitution, abstract function IBP, definite integral boundary term, product rule for integrals