Integration by Parts and Substitution
Integration by Parts and Substitution
At A-Level, you must master two powerful techniques for integrating functions that cannot be integrated directly: integration by substitution (reverse chain rule) and integration by parts (reverse product rule).
Integration by Substitution
This technique reverses the chain rule. If a function is a composite, you introduce a substitution u = g(x) to simplify the integral.
Method:
1. Choose u = g(x) — usually the "inner function" or the most awkward part
2. Find du/dx and rearrange to express dx in terms of du
3. Replace all x terms and dx with u terms and du
4. Integrate with respect to u
5. Substitute back to get the answer in terms of x
Worked Example 1: Find ∫ 2x(x² + 3)⁴ dx.
Let u = x² + 3, so du/dx = 2x, meaning du = 2x dx.
The integral becomes ∫ u⁴ du = u⁵/5 + c = (x² + 3)⁵/5 + c
Worked Example 2: Find ∫ x·√(1 + x) dx.
Let u = 1 + x, so x = u - 1 and dx = du.
∫ (u - 1)·√u du = ∫ (u^(3/2) - u^(1/2)) du
= (2/5)u^(5/2) - (2/3)u^(3/2) + c
= (2/5)(1 + x)^(5/2) - (2/3)(1 + x)^(3/2) + c
Definite Integrals with Substitution
For definite integrals, change the limits when you substitute. If u = g(x), the new limits are u₁ = g(a) and u₂ = g(b). Then you do not need to substitute back.
Worked Example: Evaluate ∫₀² x/(x² + 1) dx.
Let u = x² + 1, du = 2x dx, so x dx = du/2.
When x = 0: u = 1. When x = 2: u = 5.
∫₁⁵ (1/u)(du/2) = (1/2)[ln|u|]₁⁵ = (1/2)(ln5 - ln1) = ln5/2 ≈ 0.805
Integration by Parts
This technique reverses the product rule. The formula is:
∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx
Or in shorthand: ∫ u dv = uv - ∫ v du
Choosing u and dv: Use the LIATE rule (Logarithms, Inverse trig, Algebraic, Trig, Exponentials) — choose u from the highest item in the list.
Worked Example 1: Find ∫ x·eˣ dx.
Choose u = x (algebraic — higher in LIATE) and dv = eˣ dx.
Then du = dx and v = eˣ.
∫ xeˣ dx = xeˣ - ∫ eˣ dx = xeˣ - eˣ + c = eˣ(x - 1) + c
Worked Example 2: Find ∫ x²sinx dx.
Choose u = x², dv = sinx dx. Then du = 2x dx, v = -cosx.
∫ x²sinx dx = -x²cosx - ∫ (-cosx)(2x) dx = -x²cosx + 2∫ xcosx dx
Now apply by parts again to ∫ xcosx dx: u = x, dv = cosx dx, du = dx, v = sinx.
∫ xcosx dx = xsinx - ∫ sinx dx = xsinx + cosx
Combining: -x²cosx + 2xsinx + 2cosx + c
Special Cases
Integrating ln(x): Write as ∫ 1·lnx dx. Choose u = lnx, dv = 1 dx.
∫ lnx dx = xlnx - ∫ x·(1/x) dx = xlnx - x + c = x(lnx - 1) + c
Cyclic integrals: Sometimes integration by parts leads back to the original integral. Set up an equation and solve.
Example: Find I = ∫ eˣsinx dx.
By parts twice, you arrive at: I = eˣsinx - eˣcosx - I
2I = eˣ(sinx - cosx)
I = eˣ(sinx - cosx)/2 + c
Choosing Between Substitution and By Parts
| Integral shape | Technique |
|---|---|
| f(g(x))·g'(x) — composite with inner derivative present | Substitution |
| Product of two different types (e.g. x·eˣ, x·sinx, x·lnx) | By parts |
| Rational functions with factored denominator | Partial fractions (see next topic) |
Exam Tips
- For substitution, make sure no x terms remain after substituting — if they do, express x in terms of u.
- For definite integrals with substitution, always change the limits. This avoids the messy step of substituting back.
- By parts problems sometimes need two applications — be prepared to persist.
- Recognise standard results: ∫ lnx dx = x(lnx - 1) + c; ∫ xⁿeˣ dx requires n rounds of by parts.
- Always check your answer by differentiating it. The derivative of your result should give back the integrand.