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Integration by parts (SACE Stage 2 Specialist Mathematics)

Syllabus dot point

“Evaluate integrals using integration by parts, including choosing u and dv and applying the method more than once where needed.”

8 min read

Quick answer

Integration by parts reverses the product rule: ∫u dv=uv−∫v du\displaystyle\int u\,dv = uv - \int v\,du. Choose uu as the part that gets simpler when differentiated and dvdv as the part you can integrate. The LIATE order - logs, inverse trig, algebraic, trig, exponential - is a reliable guide for picking uu. Sometimes you apply the method twice.

Jump to a section
  1. What this dot point is asking
  2. The formula
  3. Choosing u and dv
  4. A first example
  5. Why the choice of u matters
  6. Integrating the logarithm
  7. Applying the method twice
  8. The cyclic case
  9. Exam-style questions

What this dot point is asking

You need to evaluate integrals using integration by parts, choosing uu and dvdv sensibly and applying the method repeatedly when required.

The formula

Starting from the product rule ddx(uv)=udvdx+vdudx\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx} and integrating both sides gives

Integration by parts

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

You split the integrand into a part to differentiate (uu) and a part to integrate (dvdv). The goal is that the new integral ∫v du\int v\,du is simpler than the original.

Choosing u and dv

The LIATE guide

To choose uu, prefer whichever type comes first in LIATE: Logarithmic, Inverse trig, Algebraic (powers of xx), Trigonometric, Exponential. Whatever is left becomes dvdv. This usually makes ∫v du\int v\,du simpler, because the chosen uu differentiates toward something tidier.

A first example

Integration by parts once

Find ∫xex dx\displaystyle\int x e^{x}\,dx

Step 1: Choose uu and dvdv using LIATE

LIATE lists Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential, and the type that appears first should be chosen as uu. Here the algebraic term xx ranks higher than the exponential exe^x, so we set u=xu = x and dv=ex dxdv = e^{x}\,dx. This choice works because differentiating xx reduces it to a constant, making the remaining integral easier.

Step 2: Find dudu and vv

Differentiate uu and integrate dvdv to get the pieces needed for the formula.

du=dx,v=∫ex dx=ex.du = dx, \qquad v = \int e^{x}\,dx = e^{x}.

Step 3: Apply the formula and simplify

Substituting into ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du and evaluating the remaining integral gives the answer.

∫xex dx=uv−∫v du=xex−∫ex dx=xex−ex+C=ex(x−1)+C.\int x e^{x}\,dx = uv - \int v\,du = x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} + C = e^{x}(x - 1) + C.

Final answer: ex(x−1)+Ce^{x}(x - 1) + C

Why the choice of u matters

The formula trades the original integral ∫u dv\int u\,dv for a new one ∫v du\int v\,du. The method only helps if that new integral is easier, which happens when uu becomes simpler on differentiation. A power of xx differentiates toward a lower power and eventually a constant, so it is usually a good uu when paired with an exponential or trig dvdv. A logarithm or inverse trig function, by contrast, has no elementary antiderivative of its own but a tidy derivative, so it should almost always be chosen as uu. The LIATE order encodes exactly this reasoning, which is why following it so reliably produces a simpler second integral.

Integrating the logarithm

A classic case is ∫ln⁡x dx\int \ln x\,dx, where there is seemingly no product. Take u=ln⁡xu = \ln x and dv=dxdv = dx:

∫ln⁡x dx=xln⁡x−∫x⋅1x dx=xln⁡x−∫1 dx=xln⁡x−x+C.\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.

Applying the method twice

When the algebraic factor is a higher power, one pass reduces it but leaves another integral by parts.

Applying integration by parts twice

Find ∫x2sin⁡x dx\displaystyle\int x^{2}\sin x\,dx

Step 1: First application, choose u=x2u = x^2

By LIATE the algebraic x2x^2 ranks above the trigonometric sin⁡x\sin x, so set u=x2u = x^2 and dv=sin⁡x dxdv = \sin x\,dx. Differentiating and integrating gives du=2x dxdu = 2x\,dx and v=−cos⁡xv = -\cos x.

Step 2: Apply the formula to get the remaining integral

Substituting into the integration by parts formula reduces the power of xx by one, but the result still contains a product that needs a second pass.

∫x2sin⁡x dx=−x2cos⁡x−∫(−cos⁡x)(2x) dx=−x2cos⁡x+2∫xcos⁡x dx.\int x^2\sin x\,dx = -x^2\cos x - \int (-\cos x)(2x)\,dx = -x^2\cos x + 2\int x\cos x\,dx.

Step 3: Second application, integrate ∫xcos⁡x dx\int x\cos x\,dx

Apply integration by parts again with u=xu = x and dv=cos⁡x dxdv = \cos x\,dx, giving du=dxdu = dx and v=sin⁡xv = \sin x. Now the remaining integral is a single trig function, which is straightforward.

∫xcos⁡x dx=xsin⁡x−∫sin⁡x dx=xsin⁡x+cos⁡x.\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x.

Step 4: Substitute back and collect terms

Replace ∫xcos⁡x dx\int x\cos x\,dx in the result from Step 2 with the expression found in Step 3, then simplify.

∫x2sin⁡x dx=−x2cos⁡x+2(xsin⁡x+cos⁡x)+C=−x2cos⁡x+2xsin⁡x+2cos⁡x+C.\int x^2\sin x\,dx = -x^2\cos x + 2\big(x\sin x + \cos x\big) + C = -x^2\cos x + 2x\sin x + 2\cos x + C.

Definite integrals by parts

For a definite integral, apply ∫abu dv=[uv]ab−∫abv du\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du, evaluating the boundary term [uv]ab[uv]_a^b at the limits and integrating the remaining piece between the same limits.

The cyclic case

A special situation arises with integrals such as ∫excos⁡x dx\int e^x\cos x\,dx, where applying integration by parts twice returns a multiple of the original integral. Rather than continuing forever, you set the original integral equal to the expression you have derived, treat it as an unknown, and solve algebraically. For example, two passes give I=exsin⁡x+excos⁡x−II = e^x\sin x + e^x\cos x - I, so 2I=ex(sin⁡x+cos⁡x)2I = e^x(\sin x + \cos x) and I=12ex(sin⁡x+cos⁡x)+CI = \tfrac{1}{2}e^x(\sin x + \cos x) + C. Recognising when the integral reappears, and solving for it instead of integrating again, is the trick this type of question rewards.

Common errors
Choosing uu and dvdv the wrong way round
A poor choice produces an integral harder than the original. Follow LIATE and check the new integral is simpler.
Forgetting the minus sign in the formula
It is uv−∫v duuv - \int v\,du, not plus. Dropping it flips the sign of the second term.
Sign slip when vv involves a trig function
Integrating sin⁡x\sin x gives −cos⁡x-\cos x; that minus then interacts with the formula's minus. Track signs carefully.
Losing the constant of integration
Indefinite integrals need +C+C at the end.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2022-style3 marks
Calculator-free. Use integration by parts to find ∫xe2x dx\displaystyle\int x e^{2x}\,dx.
Show worked answer →

Integration by parts: ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du.

Choose u=xu = x (so du=dxdu = dx) and dv=e2x dxdv = e^{2x}\,dx (so v=12e2xv = \tfrac{1}{2}e^{2x}), taking the algebraic factor as uu. [1 mark]

∫xe2x dx=x⋅12e2x−∫12e2x dx=12xe2x−12∫e2x dx.  [1 mark]\int x e^{2x}\,dx = x\cdot\tfrac{1}{2}e^{2x} - \int \tfrac{1}{2}e^{2x}\,dx = \tfrac{1}{2}x e^{2x} - \tfrac{1}{2}\int e^{2x}\,dx. \;[\text{1 mark}]

The remaining integral is 12e2x\tfrac{1}{2}e^{2x}, so =12xe2x−14e2x+c= \tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x} + c. [1 mark]

SACE 2023-style3 marks
Calculator-free. Given ∫xe2x dx=12xe2x−14e2x+c\displaystyle\int x e^{2x}\,dx = \tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x} + c, use integration by parts to show that ∫x2e2x dx=e2x(12x2−12x+14)+k\displaystyle\int x^2 e^{2x}\,dx = e^{2x}\left(\tfrac{1}{2}x^2 - \tfrac{1}{2}x + \tfrac{1}{4}\right) + k.
Show worked answer →

Choose u=x2u = x^2 (so du=2x dxdu = 2x\,dx) and dv=e2x dxdv = e^{2x}\,dx (so v=12e2xv = \tfrac{1}{2}e^{2x}). [1 mark]

∫x2e2x dx=12x2e2x−∫12e2x(2x) dx=12x2e2x−∫xe2x dx.  [1 mark]\int x^2 e^{2x}\,dx = \tfrac{1}{2}x^2 e^{2x} - \int \tfrac{1}{2}e^{2x}(2x)\,dx = \tfrac{1}{2}x^2 e^{2x} - \int x e^{2x}\,dx. \;[\text{1 mark}]

Substitute the given result: =12x2e2x−(12xe2x−14e2x)+k=e2x(12x2−12x+14)+k= \tfrac{1}{2}x^2 e^{2x} - \left(\tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x}\right) + k = e^{2x}\left(\tfrac{1}{2}x^2 - \tfrac{1}{2}x + \tfrac{1}{4}\right) + k. [1 mark]

SACE 2021-style2 marks
Calculator-free. Find ∫ln⁡x dx\displaystyle\int \ln x\,dx.
Show worked answer →

There is no obvious product, so take u=ln⁡xu = \ln x and dv=dxdv = dx, giving du=1x dxdu = \tfrac{1}{x}\,dx and v=xv = x:

∫ln⁡x dx=xln⁡x−∫x⋅1x dx=xln⁡x−∫1 dx=xln⁡x−x+C.\int \ln x\,dx = x\ln x - \int x\cdot\tfrac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.

Marks: one for the choice u=ln⁡xu = \ln x, dv=dxdv = dx, one for the answer xln⁡x−x+Cx\ln x - x + C.

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Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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