Unit 1
Techniques of Integration
Differentiation is an algorithm — follow the rules and you always get an answer. Integration is a search. This unit is the toolkit for that search: six techniques, each of which converts some class of unfamiliar integral into one you already recognise. The skill being graded is not any single technique. It is knowing which one to reach for.
How to read an integral before you touch it
Most people lose time on integrals by computing before they have looked. Spend ten seconds asking three questions, in this order:
- Can I simplify it algebraically? Expand a product, split a fraction into separate terms, apply a trig identity. A surprising number of frightening integrals collapse here and need no technique at all.
- Is there an inner function whose derivative is also present? That is substitution, and it is the cheapest tool available.
- What kind of object is this? A product of unlike things? A rational function? Something with a radical? Each kind has its own tool — the list is in section 8.
Differentiate your answer. Integration is hard to do and trivial to check, so the asymmetry is entirely in your favour. Ten seconds of differentiating catches sign errors, dropped constants and missing chain-rule factors — which is most of what actually goes wrong.
1 · Substitution, properly
Substitution is the chain rule read backwards. When you differentiate \(F(g(x))\) you get \(F'(g(x))\,g'(x)\) — an inner function and a copy of its derivative. So when you see that pattern inside an integrand, you are looking at something that came from a chain rule, and you can undo it.
The "derivative is also present" condition is looser than it sounds: it only has to be there up to a constant factor, because constants move freely through an integral. If \(u = x^2\) then \(du = 2x\,dx\), and an integrand containing a bare \(x\,dx\) is perfectly fine — just write \(x\,dx = \tfrac12\,du\).
You have two legal options, and one illegal one.
- Convert the limits to \(u\)-values and never look back. Usually cleanest.
- Drop the limits, integrate, back-substitute to \(x\), then evaluate.
- Never keep the original \(x\) limits on a \(u\) integral. \(\int_0^2\) does not stay \(\int_0^2\) when the variable changes meaning.
2 · Integration by parts
The product rule read backwards. Start from \((uv)' = u'v + uv'\) and integrate both sides:
Parts does not solve an integral; it trades one integral for another. The whole game is making the trade favourable: choose \(u\) so that \(du\) is simpler than \(u\), while \(dv\) is still something you can actually integrate.
The mnemonic is LIATE — pick \(u\) as whichever type appears first:
| Priority for \(u\) | Type | Example | Why it works |
|---|---|---|---|
| L | Logarithmic | \(\ln x\) | You cannot easily integrate \(\ln x\), but its derivative \(1/x\) is trivial |
| I | Inverse trig | \(\arctan x\) | Same reason — easy to differentiate, hard to integrate |
| A | Algebraic | \(x^2,\; x^3\) | Differentiating lowers the degree; repeat and it dies |
| T | Trigonometric | \(\sin x,\; \cos x\) | Neither simplifies; it is a tiebreak |
| E | Exponential | \(e^{x}\) | Easiest thing to integrate, so make it \(dv\) |
The tabular method
When \(u\) is a polynomial that will eventually differentiate to zero, do not write out three separate rounds of parts. Build two columns — differentiate the left, integrate the right — and pair them diagonally with alternating signs.
| Sign | Differentiate: \(x^3\) | Integrate: \(e^{2x}\) |
|---|---|---|
| \(+\) | \(x^3\) | \(e^{2x}\) |
| \(-\) | \(3x^2\) | \(\tfrac12 e^{2x}\) |
| \(+\) | \(6x\) | \(\tfrac14 e^{2x}\) |
| \(-\) | \(6\) | \(\tfrac18 e^{2x}\) |
| \(+\) | \(0\) stop | \(\tfrac1{16} e^{2x}\) |
Multiply along the diagonals — row \(n\) on the left times row \(n+1\) on the right:
The cyclic trick
When parts sends you in a circle — an exponential against a sine or cosine — do parts twice, notice the original integral has reappeared on the right-hand side, and then solve for it algebraically as if it were a variable. The results are worth memorising outright, because they are exactly the integrals AC circuit analysis produces:
\(e^{-t/\tau}\cos(\omega t)\) is the ringing of an underdamped RLC circuit. Integrating it — to get charge from current, or energy from power — is exactly the cyclic case above. You meet it again in Unit 6.
3 · Trigonometric integrals
These look like a pile of special cases, but one idea runs underneath all of them: use a Pythagorean identity to convert all but one factor into a single trig function, and let the factor you held back become \(du\).
Powers of sine and cosine
For \(\displaystyle\int \sin^m x\,\cos^n x\,dx\):
| Case | Move | Identity used |
|---|---|---|
| \(n\) odd (cosine odd) | Peel off one \(\cos x\) to serve as \(du\); convert the rest. Let \(u=\sin x\). | \(\cos^2 x = 1-\sin^2 x\) |
| \(m\) odd (sine odd) | Peel off one \(\sin x\); convert the rest. Let \(u=\cos x\). | \(\sin^2 x = 1-\cos^2 x\) |
| Both even | No factor to spare — knock the powers down with half-angle formulas, then look again. | \(\sin^2 x = \dfrac{1-\cos 2x}{2},\quad \cos^2 x = \dfrac{1+\cos 2x}{2}\) |
Why odd powers are the easy case: an odd power always has one factor to spare. Peel it off and what remains is an even power, which converts cleanly by Pythagoras — because the identity is written in \(\sin^2\) and \(\cos^2\).
Powers of tangent and secant
For \(\displaystyle\int \tan^m x\,\sec^n x\,dx\):
| Case | Move | Identity used |
|---|---|---|
| \(n\) even (secant even) | Peel off \(\sec^2 x\) for \(du\). Let \(u=\tan x\). | \(\sec^2 x = 1+\tan^2 x\) |
| \(m\) odd (tangent odd) | Peel off \(\sec x\tan x\) for \(du\). Let \(u=\sec x\). | \(\tan^2 x = \sec^2 x - 1\) |
| \(m\) even, \(n\) odd | The hard case. Convert to secants and use a reduction formula or repeated parts. | \(\tan^2 x = \sec^2 x - 1\) |
The pairing is not arbitrary: \(\frac{d}{dx}\tan x = \sec^2 x\) and \(\frac{d}{dx}\sec x = \sec x\tan x\). Each rule peels off exactly the derivative of the substitution it then makes.
Products of different frequencies
For \(\int \sin(mx)\cos(nx)\,dx\) and friends, convert the product into a sum with the product-to-sum identities, then integrate term by term:
Integrate \(\cos(mx)\cos(nx)\) over a full period. Product-to-sum turns it into two cosines, and both integrate to zero over a whole number of cycles — unless \(m=n\), where the \(\cos(A-B)\) term becomes \(\cos 0 = 1\) and survives:
That single fact is why Fourier series work — it is what lets you pull one harmonic out of a signal containing infinitely many. It is also why mixers, lock-in amplifiers and superheterodyne radios function.
4 · Trigonometric substitution
When a radical of the form \(\sqrt{a^2-x^2}\), \(\sqrt{a^2+x^2}\) or \(\sqrt{x^2-a^2}\) blocks every other method, trade \(x\) for a trig function chosen so that the radical collapses through a Pythagorean identity. The radical is the enemy; these substitutions are the only tool that reliably removes one.
| Radical | Substitute | \(dx\) | Radical becomes | Identity |
|---|---|---|---|---|
| \(\sqrt{a^2-x^2}\) | \(x = a\sin\theta\) | \(a\cos\theta\,d\theta\) | \(a\cos\theta\) | \(1-\sin^2 = \cos^2\) |
| \(\sqrt{a^2+x^2}\) | \(x = a\tan\theta\) | \(a\sec^2\theta\,d\theta\) | \(a\sec\theta\) | \(1+\tan^2 = \sec^2\) |
| \(\sqrt{x^2-a^2}\) | \(x = a\sec\theta\) | \(a\sec\theta\tan\theta\,d\theta\) | \(a\tan\theta\) | \(\sec^2-1 = \tan^2\) |
Always draw the triangle. You have to convert back from \(\theta\) to \(x\) at the end, and the triangle hands you every trig function of \(\theta\) with no algebra at all — just read off the sides.
A radical like \(\sqrt{x^2+6x+13}\) matches none of the three cases — until you complete the square:
Now it is the tangent case with \(a=2\) and a shifted variable. Always complete the square before concluding that trig substitution does not apply. The same move rescues many partial-fraction and arctangent problems.
5 · Partial fractions
For a rational function \(P(x)/Q(x)\), split it into a sum of pieces simple enough to integrate. This is the reverse of putting fractions over a common denominator, and it is the single highest-leverage technique in this unit for an engineer.
Two preconditions, in this order:
- Degree check. If \(\deg P \ge \deg Q\), do polynomial long division first. Partial fractions only applies to a proper rational function.
- Factor \(Q\) completely into linear and irreducible quadratic factors.
Then assign one term per factor according to its type:
| Factor in \(Q(x)\) | Contributes |
|---|---|
| Distinct linear, \((x-r)\) | \(\dfrac{A}{x-r}\) |
| Repeated linear, \((x-r)^k\) | \(\dfrac{A_1}{x-r} + \dfrac{A_2}{(x-r)^2} + \cdots + \dfrac{A_k}{(x-r)^k}\) |
| Irreducible quadratic, \(x^2+bx+c\) | \(\dfrac{Ax+B}{x^2+bx+c}\) (linear numerator!) |
| Repeated quadratic, \((x^2+bx+c)^k\) | One such term for each power \(1\) through \(k\) |
The cover-up method
For distinct linear factors you can skip the system of equations entirely. To find the coefficient over \((x-r)\): cover that factor in the denominator and evaluate everything left at \(x=r\).
\(B\): cover \((s+2)\), evaluate \(\dfrac{s+3}{s+1}\) at \(s=-2 \Rightarrow \dfrac{1}{-1} = -1\).
Two seconds, no algebra.
That was not a random example. In Signals & Systems you will meet \(F(s) = \dfrac{s+3}{(s+1)(s+2)}\) and need \(f(t)\). Split it by cover-up, then read each term off using \(\dfrac{1}{s+a} \longrightarrow e^{-at}\):
You will run partial fractions more times in your degree than every other integration technique combined. Repeated roots become \(te^{-at}\) terms; complex conjugate roots become damped sinusoids. Those three cases are exactly the three cases of second-order ODEs — the same algebra approached from opposite ends.
6 · Improper integrals
An integral is improper if the interval is infinite (Type I) or the integrand blows up somewhere on it (Type II). In both cases the fix is the same: replace the bad point with a limit, then do ordinary calculus.
\(\displaystyle\int_{-1}^{1}\frac{dx}{x^2}\) looks perfectly ordinary and evaluates naively to \(-2\) — a negative answer for a strictly positive integrand, which is impossible. The integrand blows up at \(x=0\), inside the interval. Split there and take limits from both sides; the integral diverges.
Always scan for infinite discontinuities inside the limits, not just at the ends. A nonsensical sign is often the only warning you get.
The p-integrals — memorise both
| Integral | Converges when | Value | Intuition |
|---|---|---|---|
| \(\displaystyle\int_1^{\infty}\frac{dx}{x^p}\) | \(p > 1\) | \(\dfrac{1}{p-1}\) | The tail must decay fast for the area out to infinity to be finite |
| \(\displaystyle\int_0^{1}\frac{dx}{x^p}\) | \(p < 1\) | \(\dfrac{1}{1-p}\) | The spike must be mild for the area near zero to be finite |
Comparison test
When you cannot evaluate an improper integral you can often still decide whether it converges. For \(0 \le f \le g\) on \([a,\infty)\):
- If \(\int g\) converges, so does \(\int f\). (Smaller than something finite.)
- If \(\int f\) diverges, so does \(\int g\). (Bigger than something infinite.)
The other two combinations tell you nothing — being smaller than something infinite is no information at all. This is the same logic as the comparison test for series in Unit 3; the integral test is precisely the bridge between them.
The Laplace transform is an improper integral:
The "region of convergence" you will hear so much about is nothing more than the set of \(s\) for which this improper integral converges. Same story for the Fourier transform, and for total signal energy \(\int_{-\infty}^{\infty}|x(t)|^2\,dt\).
7 · Numerical integration
Most integrals that arise in practice have no closed form — including ones as innocent-looking as \(\int e^{-x^2}dx\). And when your data comes from an oscilloscope rather than an equation, there is no formula to integrate at all. You approximate.
| Rule | Formula, with \(\Delta x = \frac{b-a}{n}\) | Error bound | Exact for |
|---|---|---|---|
| Midpoint | \(\Delta x\big[f(\bar x_1) + \cdots + f(\bar x_n)\big]\) | \(\dfrac{K(b-a)^3}{24n^2}\) | Linear |
| Trapezoid | \(\dfrac{\Delta x}{2}\big[f_0 + 2f_1 + \cdots + 2f_{n-1} + f_n\big]\) | \(\dfrac{K(b-a)^3}{12n^2}\) | Linear |
| Simpson (\(n\) must be even) |
\(\dfrac{\Delta x}{3}\big[f_0 + 4f_1 + 2f_2 + 4f_3 + \cdots + 4f_{n-1} + f_n\big]\) | \(\dfrac{K(b-a)^5}{180n^4}\) | Cubics |
Doubling \(n\) cuts trapezoid error by 4× and Simpson error by 16×. That gap is why Simpson is the practical default, and why the numerical integrators inside SPICE and MATLAB are built on higher-order rules rather than trapezoids.
8 · Choosing a technique
Under exam pressure, run this list from the top and stop at the first line that applies. Do not start with the exotic techniques — most integrals fall to the first two lines.
- Simplify algebraically first. Expand, split a fraction into separate terms, use a trig identity.
- Is there an inner function whose derivative is present? → substitution.
- A product of two unlike things (polynomial × exponential, polynomial × log, exponential × trig)? → parts, choosing \(u\) by LIATE.
- Only trig functions? → trig-integral rules; check the odd/even powers.
- A rational function? → degree check, factor, partial fractions.
- A radical of the form \(a^2\pm x^2\) or \(x^2-a^2\)? → complete the square if needed, then trig substitution.
- Still stuck? Try a substitution that rationalises the radical, or manipulate the integrand into something on this list. Then consider that it may have no elementary antiderivative — go numerical.
9 · Worked examples
1. Parts, twice — \(\int x^2 e^{-3x}\,dx\)
Polynomial times exponential, so LIATE says \(u = x^2\) (A before E). Use the table:
| Sign | D | I |
|---|---|---|
| \(+\) | \(x^2\) | \(e^{-3x}\) |
| \(-\) | \(2x\) | \(-\tfrac13 e^{-3x}\) |
| \(+\) | \(2\) | \(\tfrac19 e^{-3x}\) |
| \(-\) | \(0\) | \(-\tfrac1{27} e^{-3x}\) |
Check: differentiating the factored form returns \(x^2 e^{-3x}\). ✓
2. Odd power of sine — \(\int \sin^3 x\,\cos^2 x\,dx\)
Sine carries the odd power, so peel one off and let \(u=\cos x\), \(du = -\sin x\,dx\).
The key recognition: an odd power always leaves one factor to donate to \(du\), and what remains is even, so Pythagoras converts it cleanly.
3. Trig substitution — \(\int \dfrac{dx}{x^2\sqrt{x^2+4}}\)
Form \(\sqrt{a^2+x^2}\) with \(a=2\), so \(x = 2\tan\theta\), \(dx = 2\sec^2\theta\,d\theta\), \(\sqrt{x^2+4} = 2\sec\theta\).
Convert to sines and cosines — almost always the right move at this stage:
Now substitute \(w = \sin\theta\):
Back-substitute from the tangent triangle, where \(\sin\theta = x/\sqrt{x^2+4}\):
4. Partial fractions with a repeated factor
Evaluate \(\displaystyle\int \frac{3x+5}{(x-1)(x+2)^2}\,dx\).
Proper already (degree 1 over degree 3). One repeated factor, so:
Multiplying through: \(3x+5 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)\).
Strategic values first. These two come free:
- \(x = 1\): \(8 = 9A \Rightarrow A = \tfrac89\)
- \(x = -2\): \(-1 = -3C \Rightarrow C = \tfrac13\)
Then one more equation for \(B\). Compare \(x^2\) coefficients: \(0 = A + B\), so \(B = -\tfrac89\).
Note the pattern: simple factors give logarithms, the repeated factor gives a power. In Laplace terms that repeated root produces a \(te^{-at}\) term instead of a plain exponential — the signature of critical damping.
5. Improper integral — deriving a Laplace transform
Show that \(\displaystyle\int_0^{\infty} e^{-at}e^{-st}\,dt = \frac{1}{s+a}\), and state when it converges.
Combine the exponentials and write the limit properly:
The first term \(\to 0\) if and only if \(s+a > 0\). Under that condition:
If \(s+a \le 0\) the exponential grows without bound and the integral diverges: the transform simply does not exist there.
6. Simpson’s rule on measured data
A current \(i(t)\) is sampled every 0.5 s. Estimate the charge delivered, \(q = \int_0^2 i\,dt\).
| \(t\) (s) | 0.0 | 0.5 | 1.0 | 1.5 | 2.0 |
|---|---|---|---|---|---|
| \(i\) (A) | 0.00 | 1.85 | 2.60 | 2.15 | 1.20 |
\(n = 4\) intervals (even ✓), \(\Delta t = 0.5\). Weights 1, 4, 2, 4, 1:
Trapezoid on the same data gives 3.64 C. With no formula for \(i(t)\) there is nothing to integrate symbolically — this is the normal situation once you are working with real instruments.
Checklist
- I run the strategy list from the top instead of guessing a technique.
- I choose \(u\) for parts by LIATE, and I use the table when the polynomial will die.
- I know the two cyclic \(e^{ax}\)-times-trig results by heart.
- I check odd/even powers before touching a trig integral.
- I complete the square before ruling out trig substitution, and I draw the triangle.
- I do the degree check before partial fractions, and I use cover-up for distinct linear factors.
- I write improper integrals as limits, and I check for singularities inside the interval.
- I differentiate my answer to check it.