How to integrate by parts
How to choose u so the integral gets simpler, a worked example, and what to do when the same integral reappears on the right.
What the examiner is testing
Whether you can choose \( u \) sensibly. The formula is given in the booklet; the marks are for making the integral on the right-hand side simpler than the one you started with.
$$ \int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx $$
Choosing u — the LIATE order
Take \( u \) to be whichever of these appears first in the list:
- Logarithmic — \( \ln x \)
- Inverse trigonometric — \( \arctan x \)
- Algebraic — \( x^2 \), \( 3x \)
- Trigonometric — \( \sin x \), \( \cos x \)
- Exponential — \( e^{2x} \)
The logic is simply that \( \ln x \) and powers of \( x \) get simpler when differentiated, while \( e^x \) and \( \sin x \) never do.
The method
- Choose \( u \) by LIATE; everything else is \( \dfrac{dv}{dx} \).
- Differentiate \( u \), integrate \( \dfrac{dv}{dx} \). Write all four pieces out.
- Substitute into the formula.
- Evaluate the new integral — it should be easier.
- Add \( + c \), or apply the limits to both parts if it is definite.
Worked example
Find \( \displaystyle\int x\cos x \, dx \).
Step 1 — choose. Algebraic beats trigonometric in LIATE, so \( u = x \).
Step 2 — the four pieces.
$$ u = x \quad \Rightarrow \quad \frac{du}{dx} = 1 $$
$$ \frac{dv}{dx} = \cos x \quad \Rightarrow \quad v = \sin x $$
Step 3 — substitute.
$$ \int x\cos x \, dx = x\sin x - \int \sin x \times 1 \, dx $$
Step 4 — evaluate. The remaining integral is now a standard one:
$$ = x\sin x - (-\cos x) = x\sin x + \cos x $$
Step 5 — constant.
$$ \int x\cos x \, dx = x\sin x + \cos x + c $$
Check by differentiating: the product rule on \( x\sin x \) gives \( \sin x + x\cos x \), and differentiating \( \cos x \) gives \( -\sin x \). They sum to \( x\cos x \). Correct.
If you had chosen wrongly
Taking \( u = \cos x \) would have produced \( \int \frac{x^2}{2}\sin x \, dx \) — a higher power than you started with. That is the signal to stop and swap.
The three mistakes that lose marks
1. Choosing \( u \) to be the exponential or the trig function when an algebraic factor is present. The power climbs and the integral never terminates.
2. Losing the minus sign in front of the second integral. The formula is \( uv - \int v\,du \). Write the minus before you substitute anything.
3. Applying limits to only one part. In a definite integral, the limits apply to the \( uv \) term as well: \( \left[uv\right]_a^b - \int_a^b v\,du \).
30-second recap
LIATE picks \( u \). Write out all four pieces before substituting. If the new integral is harder, you chose backwards. Differentiate your answer to check — it takes seconds and catches every sign error.