1The idea in one paragraph

Integration has no product rule. Instead you have two ways of turning an integral you cannot do into one you can. Integration by substitution renames part of the integrand as a new variable, u, so that the integral becomes a standard one in u. It is the chain rule read backwards. Integration by parts swaps ∫ u (dv/dx) dx for uv − ∫ v (du/dx) dx. It is the product rule read backwards, and it works when u is the factor that gets simpler as you differentiate it. Every question here comes down to one decision: which technique, and which u.

2Substitution: what it does and why it works

At SL you met ∫ 2x cos(x²) dx. The inside is x², its derivative 2x sits outside, and the answer is sin(x²) + C by inspection. As a substitution, the working looks like this.

The step du = 2x dx says a small change in x makes a change in u that is 2x times as big. Figure 1 shows the definite version from 0 to 1: the area under y = 2x cos(x²) on the left, and after u = x² the area under y = cos u on the right. The curves look nothing alike, but the substitution stretches the strips so that each keeps its area, and both totals are sin 1 ≈ 0.841.

The guide describes this SL form as ∫ k g′(x) f(g(x)) dx: a function of an inside function g(x), multiplied by a constant times the derivative of that inside. When an integral has this shape, no substitution will be given, and you are expected to spot it: in ∫ 6x²(x³ − 2)⁵ dx the inside x³ − 2 has derivative 3x², and 6x² is twice that, so the answer is 2 × (x³ − 2)⁶/6 + C = (x³ − 2)⁶/3 + C.

At HL the derivative of the inside is not quite there, or leftover x terms must be rewritten in u. For those, the guide says the substitution will be provided, so the marks are for carrying it out cleanly.

3Using a substitution you are given

The method is always the same four moves.

Substitution: (1) differentiate the substitution to link du and dx; (2) replace every x and the dx, so that no x is left; (3) integrate in u; (4) put x back (or, for a definite integral, change the limits instead).

Move (2) is where most students stop too soon: every x must go, not only the "inside".

Example 1. Use the substitution u = x + 3 to find ∫ x√(x + 3) dx.

The inside is x + 3 and its derivative is 1, so dx = du. But there is a spare x outside the root. Rearrange the substitution to deal with it: x = u − 3.

Differentiating the final answer brings back x√(x + 3), which is how you check it on Paper 1.

Example 2. Use the substitution u = eˣ + 1 to find ∫ e^2x ÷ (eˣ + 1) dx.

Here du = eˣ dx. The numerator e^2x is eˣ × eˣ, so one eˣ goes with the dx to make du, and the other must be written in u: eˣ = u − 1.

No modulus is needed on ln u, because eˣ + 1 is always positive.

4Definite integrals: change the limits

For a definite integral you can put x back and use the original limits, or change the limits into u-values and never go back to x. The second is shorter, and it is what a mark scheme expects after "use the substitution … to find the exact value".

Example 3. Use the substitution u = 2x + 1 to find the exact value of ∫₀⁴ x ÷ √(2x + 1) dx.

Once the limits are in u, never substitute x = 4 into an expression in u. That mixed step is the most common way to lose the final A1 here.

Example 4: a substitution that replaces x itself. Sometimes the question gives x as a function of the new variable. The classic case is √(a² − x²), which x = a sin θ turns into a clean cosine. Find the exact value of ∫₀¹ √(1 − x²) dx using x = sin θ, for 0 ≤ θ ≤ π/2.

The restriction on θ is what makes √(cos²θ) equal cos θ rather than |cos θ|. The curve y = √(1 − x²) for 0 ≤ x ≤ 1 is a quarter of the unit circle, whose area is π/4, and Figure 6(a) in section 7 draws it. For the indefinite integral you would return to x with θ = arcsin x and sin 2θ = 2x√(1 − x²).

5Integration by parts: where the formula comes from

Take two functions u and v of x. The product rule says

Integrate both sides with respect to x. The left side integrates back to uv. Rearranging gives the rule, which is in the formula booklet:

∫ u (dv/dx) dx = uv − ∫ v (du/dx) dx · integration by parts. You choose which factor is u (you will differentiate it) and which is dv/dx (you will integrate it).

The formula is a trade: the integral on the left for the one on the right. It is worth making only if the new integral, ∫ v (du/dx) dx, is easier. Figure 2 shows the same fact as areas. Plot v against u: the area under the curve is ∫ v du, the area to its left is ∫ u dv, and together they make the large rectangle u₂v₂ minus the small one u₁v₁. Know one area and you know the other.

Example 5. Find ∫ x sin x dx.

Two different kinds of function are multiplied: a polynomial x and a trig function sin x. Differentiate x and it becomes 1, which is simpler. So u = x and dv/dx = sin x.

Differentiating gives x sin x back. Choose the other way round, u = sin x and dv/dx = x, and the new integral is ∫ (x²/2) cos x dx, with a higher power of x than you started with. When the trade makes things worse, you have chosen u badly. Stop and swap.

6Choosing u, and the three integrals that need a hidden 1

Figure 3 gives an order of preference that works for every integral in this course. Read it from the left: whichever factor appears furthest left should be u.

The logic: logarithms and inverse trig functions have simple derivatives but awkward integrals, so they must be u. Polynomials shrink when differentiated, so they make a good u against trig and exponential functions, which do not grow when integrated.

Example 6. Find ∫ x² ln x dx.

A logarithm beats a polynomial, so u = ln x.

The 1/x from ln x cancels a power of x, which is why ln x must be u.

Example 7: ln x on its own. With only one factor, write ∫ ln x dx as ∫ 1 × ln x dx, with u = ln x and dv/dx = 1, so v = x.

This result is not in the formula booklet, so be able to derive it in three lines. Figure 6(b) uses it: the area under y = ln x from 1 to e is (e − e) − (0 − 1) = 1.

Example 8: arcsin x on its own. The same hidden 1 works: u = arcsin x and dv/dx = 1.

The integral left over has the SL form: the inside 1 − x² has derivative −2x, so ∫ x(1 − x²)^−1/2 dx = −(1 − x²)^1/2 + C, and

Parts, then inspection: ∫ arctan x dx works the same way and gives x arctan x − ½ ln(1 + x²) + C.

Definite integrals by parts. Evaluate the uv part between the limits too. For ∫₀^π/2 x cos x dx, with u = x and v = sin x:

7Repeated integration by parts

When one round leaves an integral that still needs parts, there are two patterns.

Pattern 1: the polynomial runs out. In ∫ x² eˣ dx, take u = x². The new integral ∫ 2x eˣ dx is still a product, but the power has dropped to 1, and one more round finishes it.

The minus sign in front of the bracket applies to both terms of the second round, which is why −2eˣ becomes +2eˣ. Figure 4 lays the same calculation out as a table: differentiate down the left until it reaches 0, integrate down the right, and multiply along the arrows with alternating signs. Use it to organise and check, but write at least the first round in the formula's form, because that line carries the method mark.

Pattern 2: the integral comes back. In ∫ eˣ sin x dx neither factor gets simpler, and two rounds of parts bring back the integral you started with. That is not a dead end. Call it I and solve for it like an unknown.

In the cyclic case, keep the same type of function as u in both rounds (trig both times, or exponential both times). Switch, and the second round undoes the first, leaving the useless line I = I.

Figure 5 shows the area under y = eˣ sin x from 0 to π:

Figure 6 checks two earlier results with pictures: π/4 from Example 4 and ∫₁^e ln x dx = 1 from Example 7. On Paper 2 a GDC's numerical integral does the same in seconds.

8Which technique? A decision in three questions

Ask, in this order:

1. Is it on the standard list, or a composite with ax + b? Integrate directly (5.10, 5.15).
2. Is it a function of an inside function, times (a multiple of) the inside's derivative? Inspection, no hint given: ∫ x e^x² dx and ∫ (ln x)/x dx are this type.
3. Is it a product of two different kinds of function, or a lone ln or inverse trig? Parts, with u chosen from Figure 3.

A given substitution must be used. Some integrals need both techniques: u = x² turns ∫ x³ e^x² dx into ½ ∫ u e^u du, which then needs parts (Try it Q5). A rational function whose denominator factorises may need partial fractions (1.11, 5.15) first.