Integration Techniques — Maths for Thermodynamics & Kinetics

Two ways through a hard integral

Substitution and integration by parts

Most integrals you encounter in chemistry cannot be done by applying the power rule alone — the integrand is either a composite function (a function inside a function) or a product of unlike functions. Two techniques handle virtually every case at undergraduate level.

u-substitution undoes the chain rule. You spot an inner function $u (x)$ whose derivative already appears alongside it in the integrand; replacing $x$ with $u$ (and $d x$ with $d u / u^{'}$ ) converts a difficult integral into a standard one.

Integration by parts undoes the product rule. Its formula, $\int u d v = uv - \int v d u$ , trades the original integral for a (hopefully simpler) one involving the derivative of $u$ and the antiderivative of $d v$ . In quantum chemistry this is the engine behind expectation value integrals such as $⟨ x ⟩ = \int_{0}^{L} ψ^{*} x ψ d x$ .

Key formulas at a glance

u-substitution

\int f (g (x)) g^{'} (x) d x = \int f (u) d u

Integration by parts

\int u d v = uv - \int v d u

LIATE rule for by-parts

Choose u as: Logs, Inverse-trig, Algebraic,Trig, Exponential — whichever appears first in that list.

Standard result from sub

\int e^{- k x} d x = - \frac{1}{k} e^{- k x} + C

(let

u = - k x

)

Why chemists need them

Integrating rate laws, evaluating expectation values like

⟨ x ⟩ = \int ψ^{*} x ψ d x

, and the Gaussian integrals that underpin partition functions all require these moves. Open the technique lab and watch the area accumulate as you drag the upper limit.

Changing limits on substitution

When

u = g (x)

, update the limits too:

\int_{x = a}^{x = b} f (g) g^{'} d x = \int_{u = g (a)}^{u = g (b)} f (u) d u

LIATE priority for by-parts

Pick u first in order: Logs > Inverse-trig > Algebraic > Trig > Exponential. The remaining factor is dv.

By-parts applied twice

\int x^{2} e^{x} d x

requires two rounds; each pass lowers the power of

x

by one until the integral is standard.

∫ sin²(nπx/L) dx trick

Use

sin^{2} θ = \frac{1}{2} (1 - cos 2 θ)

to split into easy integrals. Over a full period the cosine term integrates to zero, giving

L /2

Common pitfalls

Forgetting to change limits on substitution. If you substitute $u = g (x)$ in a definite integral, the limits must change to $g (a)$ and $g (b)$ — or you must convert back to $x$ before evaluating.
Dropping the constant of integration in indefinite integrals. When using by-parts, $\int v d u$ also generates a constant; collect them all into a single $C$ at the end.
Wrong u/dv choice. Choosing $u = e^{x}$ and $d v = x d x$ for $\int x e^{x} d x$ makes $\int v d u$ harder, not easier. LIATE guides you to take $u = x$ (algebraic) instead.
Applying the power rule to n = −1. $\int x^{- 1} d x \neq = x^{0} /0$ — the antiderivative is $ln ∣ x ∣$ , not a power.

Technique lab

Switch between u-substitution and integration by parts — see the steps and drag the upper limit to accumulate the definite integral.

Technique:

\int \frac{x}{1 + x ^{2}} d x = \frac{1}{2} ln (1 + x^{2}) + C

MethodReverses the chain rule

Step 1Let u = 1 + x²

Step 2then du = 2x dx, so x dx = ½ du

Step 3∫ ½ du/u = ½ ln|u| = ½ ln(1+x²)

upper limit b2.50

definite integral 0→b0.9905

Worked example 1u-substitution on a first-order rate law

The first-order rate law gives $\frac{d [ A ]}{[ A ]} = - k d t$ . Integrate both sides from $[A]_{0}$ to $[A]$ and from $0$ to $t$ , using the substitution $u = [A]$ , to derive the integrated rate law.

Worked example 2Integration by parts for ⟨x⟩ in the particle-in-a-box

For the ground state $ψ_{1} = 2/ L sin (π x / L)$ on $[0, L]$ , evaluate the expectation value $⟨ x ⟩ = \int_{0}^{L} ψ_{1}^{*} x ψ_{1} d x$ and show it equals $L /2$ .

Worked examples

Pull both techniques together — try each before revealing the solution.

Worked example 3u-substitution: isothermal work integral

Derive the work formula for a reversible isothermal expansion using u-substitution: $W = \int_{V_{1}}^{V_{2}} \frac{n RT}{V} d V$ .

Worked example 4Integration by parts for ⟨x²⟩

Using the ground-state particle-in-a-box wavefunction, show that $⟨ x^{2} ⟩ = L^{2} (\frac{1}{3} - \frac{1}{2 π ^{2}})$ .

Worked example 5Standard Gaussian from a substitution

Partition functions for molecular translations require $\int_{0}^{\infty} e^{- α x^{2}} d x$ . Using the known result $\int_{0}^{\infty} e^{- u^{2}} d u = π /2$ and the substitution $u = α x$ , show that $\int_{0}^{\infty} e^{- α x^{2}} d x = \frac{1}{2} \frac{π}{α}$ .

Worked example 6u-substitution with limit change: Beer–Lambert absorption

The fraction of light transmitted through a pathlength $L$ in an absorbing medium satisfies $d I / d x = - α I$ . Evaluate the definite integral $\int_{I_{0}}^{I} \frac{d I}{I} = - α \int_{0}^{L} d x$ using the substitution $u = I$ , being careful to change the limits.

Worked example 7Integration by parts applied twice for ⟨x²⟩ — full derivation

Using the ground-state particle-in-a-box wavefunction $ψ_{1} = 2/ L sin (π x / L)$ , carry out the full by-parts calculation to show $⟨ x^{2} ⟩ = L^{2} (\frac{1}{3} - \frac{1}{2 π ^{2}})$ .

ChallengeChallenge — ∫ sin²(nπx/L) dx over [0, L] from first principles

Without using any look-up table, evaluate $\int_{0}^{L} sin^{2} \frac{nπ x}{L} d x$ for positive integer $n$ . Use this result to confirm the normalisation constant $N = 2/ L$ for the nth particle-in-a-box state.

Check yourself

Four quick questions on when and how to apply each technique.

Question 1 of 4 · Score 0

u-substitution is the reverse of which differentiation rule?

Choose an answer.