How does integration by parts reverse the product rule, and how do we choose the parts to evaluate products such as or ?
Apply integration by parts to evaluate integrals of products, including repeated application and the recovery of the original integral
A focused answer to the HSC Maths Extension 2 dot point on integration by parts. The formula, choosing the parts with LIATE, repeated application, and the boomerang case where the original integral reappears, with verified worked examples.
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What this dot point is asking
NESA wants you to integrate products of functions using integration by parts, the reverse of the product rule. You must choose the parts sensibly, apply the method more than once where needed, and handle the case where the original integral reappears and can be solved algebraically.
The formula
The product rule says . Integrating both sides and rearranging gives integration by parts:
The aim is to swap the original integral for an easier one. Success depends entirely on a good split: should become simpler when differentiated, and should be something you can integrate.
Choosing the parts: LIATE
A reliable order of preference for choosing is LIATE:
- Logarithmic ()
- Inverse trigonometric ()
- Algebraic ()
- Trigonometric (, )
- Exponential ()
Whatever appears earlier in this list is usually the better choice for . For example, in , the logarithm ranks above the algebraic term, so .
A first example
Evaluate . Let (algebraic) and . Then and . So
Repeated application
When the algebraic factor has a higher power, apply the method repeatedly. For , first take , to get , then reuse the result above. Each application lowers the power of by one until it vanishes.
The boomerang (recovery) case
For integrals such as , two applications of integration by parts return a multiple of the original integral. You then solve algebraically. This is the "boomerang" because the integral comes back to you, letting you collect like terms and divide.
Exam-style practice questions
Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 HSC2 marksFind integral of x e^x dx.Show worked answer →
Use integration by parts, integral of u v' dx = u v - integral of u' v dx.
Choose u = x (so u' = 1) and v' = e^x (so v = e^x).
integral of x e^x dx = x e^x - integral of (1)(e^x) dx = x e^x - e^x + C.
So integral of x e^x dx = x e^x - e^x + C, which can be written (x - 1) e^x + C.
Mark notes: 1 mark for setting up integration by parts with a sensible choice of u and v', 1 mark for the correct primitive x e^x - e^x + C.
2021 HSC1 marksWhich expression is equal to integral of x^5 e^(7x) dx? A. (1/7) x^5 e^(7x) - (5/7) integral of x^4 e^(7x) dx. B. (1/7) x^5 e^(7x) - (5/7) integral of x^5 e^(7x) dx. C. (5/7) x^4 e^(7x) - (5/7) integral of x^4 e^(7x) dx. D. (5/7) x^4 e^(7x) - (5/7) integral of x^5 e^(7x) dx.Show worked answer →
Apply one step of integration by parts to integral of x^5 e^(7x) dx.
Take u = x^5 (so u' = 5x^4) and v' = e^(7x) (so v = (1/7) e^(7x)).
integral of u v' dx = u v - integral of u' v dx
= x^5 . (1/7) e^(7x) - integral of 5x^4 . (1/7) e^(7x) dx
= (1/7) x^5 e^(7x) - (5/7) integral of x^4 e^(7x) dx.
This matches option A. The other options either differentiate instead of integrate e^(7x) (giving the wrong leading term) or leave x^5 inside the remaining integral.
The answer is A.
2023 HSC3 marksLet J_n = integral from 0 to pi/2 of sin^n q dq, where n >= 0 is an integer. Show that J_n = ((n - 1)/n) J_(n-2) for all integers n >= 2.Show worked answer →
Split off one factor of sin q and integrate by parts: write sin^n q = sin q . sin^(n-1) q.
Take u = sin^(n-1) q (so u' = (n - 1) sin^(n-2) q cos q) and v' = sin q (so v = -cos q).
J_n = [-cos q . sin^(n-1) q] from 0 to pi/2 + (n - 1) integral from 0 to pi/2 of cos^2 q sin^(n-2) q dq.
The boundary term is 0 at both limits (cos(pi/2) = 0 and sin 0 = 0 for n >= 2). So
J_n = (n - 1) integral from 0 to pi/2 of cos^2 q sin^(n-2) q dq.
Replace cos^2 q = 1 - sin^2 q:
J_n = (n - 1) integral of sin^(n-2) q dq - (n - 1) integral of sin^n q dq = (n - 1) J_(n-2) - (n - 1) J_n.
The original integral has reappeared. Collect J_n terms:
J_n + (n - 1) J_n = (n - 1) J_(n-2), so n J_n = (n - 1) J_(n-2), giving J_n = ((n - 1)/n) J_(n-2).
Mark notes: 1 mark for integration by parts with the boundary term shown to vanish, 1 mark for using cos^2 q = 1 - sin^2 q to recover J_n, 1 mark for collecting terms to reach the reduction formula.