11 min readPathway sheet3-mark and 5-mark questions; ∫ eˣ [f(x) + f′(x)] dx is a 2-mark favourite; also a JEE favourite

Integration by parts

For a product of two functions, ∫ u v dx = u ∫ v dx − ∫ [u′ ∫ v dx] dx. Choose the first function u using ILATE.

Rule∫ u v dx = u ∫ v dx − ∫ [(du/dx) ∫ v dx] dx, where u is the first function and v the second

ILATE (a memory aid for the first function)Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential

Special form∫ eˣ [f(x) + f′(x)] dx = eˣ f(x) + C

ConstantDo not add a constant when you find ∫ v dx inside the rule; add one C at the end

Where the rule comes from

It is the product rule of differentiation, read backwards.

1
Product rule: d/dx [u w] = u (dw/dx) + w (du/dx).
2
Integrate both sides: u w = ∫ u (dw/dx) dx + ∫ w (du/dx) dx.
3
Rearrange: ∫ u (dw/dx) dx = u w − ∫ w (du/dx) dx.
4
Let dw/dx = v, so w = ∫ v dx.
5
So, ∫ u v dx = u ∫ v dx − ∫ [(du/dx) ∫ v dx] dx.

Choosing the first function

Your textbook's guide: a power of x or a polynomial is usually the first function, but an inverse trigonometric or log function is taken first. The memory aid ILATE (not textbook wording) orders this: Inverse, Log, Algebraic, Trig, Exponential.
The second function v must be one you can integrate easily.
For ∫ log x dx or ∫ tan⁻¹ x dx, take 1 as the second function: ∫ log x dx = x log x − x + C.
Some integrals come back to the start, such as ∫ eˣ cos 2x dx. Apply the rule twice, call the integral I, and solve for I.
If a step makes the integral harder, you chose the wrong first function. Swap them.

Example 1: log times a power

Problem

Find ∫ x³ log x dx.

Working

A log function is taken as the first function (ILATE: L before A). So u = log x and v = x³.
∫ x³ dx = x⁴/4.
I = log x × x⁴/4 − ∫ (1/x)(x⁴/4) dx.
= (x⁴/4) log x − (1/4) ∫ x³ dx.
= (x⁴/4) log x − x⁴/16 + C.

Answer

(x⁴/4) log x − x⁴/16 + C

Example 2: the eˣ [f + f′] form

Problem

Find ∫ eˣ (x + 2)/(x + 3)² dx.

Working

Write x + 2 = (x + 3) − 1.
So (x + 2)/(x + 3)² = 1/(x + 3) − 1/(x + 3)².
Let f(x) = 1/(x + 3). Then f′(x) = −1/(x + 3)².
The integral has the form ∫ eˣ [f(x) + f′(x)] dx = eˣ f(x) + C.

Answer

eˣ/(x + 3) + C

Example 3: an integral that comes back

Problem

Find ∫ eˣ cos 2x dx.

Working

Let I = ∫ eˣ cos 2x dx. Take cos 2x as the first function (T before E).
I = cos 2x × eˣ − ∫ (−2 sin 2x) eˣ dx = eˣ cos 2x + 2 ∫ eˣ sin 2x dx.
Again by parts: ∫ eˣ sin 2x dx = eˣ sin 2x − ∫ 2 cos 2x × eˣ dx = eˣ sin 2x − 2I.
So I = eˣ cos 2x + 2(eˣ sin 2x − 2I) = eˣ cos 2x + 2eˣ sin 2x − 4I.
5I = eˣ (cos 2x + 2 sin 2x).

Answer

I = (eˣ/5)(cos 2x + 2 sin 2x) + C

In real life

Music apps and ECG machines break a signal into sine waves, called a Fourier series. Finding the size of each wave needs integrals like ∫ t sin t dt, which are done by parts.

Exam angle

2 marks: “Find ∫ eˣ (tan x + sec² x) dx.” Take f(x) = tan x, so f′(x) = sec² x. By ∫ eˣ [f + f′] dx = eˣ f + C, the answer is eˣ tan x + C.

3 marks: “Find ∫ x² e²ˣ dx.” By parts with u = x²: (x²/2) e²ˣ − ∫ x e²ˣ dx. Again by parts: ∫ x e²ˣ dx = (x/2) e²ˣ − e²ˣ/4. Answer: e²ˣ (x²/2 − x/2 + 1/4) + C.

3 marks: “Find ∫ x² (log x)² dx.” Take u = (log x)², v = x²: (x³/3)(log x)² − (2/3) ∫ x² log x dx. ∫ x² log x dx = (x³/3) log x − x³/9. Answer: (x³/3)(log x)² − (2x³/9) log x + 2x³/27 + C.

Remember it

ILATE picks u. Then: first × integral of second, minus integral of (derivative of first × integral of second).