Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 3: Pure Mathematics 3 · official syllabus
INTEGRATION LAB
Learn it by playing it
Four short missions on reversing the chain rule, substitution, parts and the double-angle trick. In each one you compute the shaded area and slide your answer onto the exact bar - when a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Reverse the chain
Differentiating e^(2x) multiplies by 2, so integrating must divide by 2: the antiderivative of e^(2x) is e^(2x)/2. The white curve is e^(2x) and the shaded region runs from 0 to 0.5. The dashed bar is the exact area; the solid bar is yours. Slide your area onto it.
Before you slide: int e^(2x) dx equals...
Your area 0.50 against the true 0.86 - remember the chain rule in reverse divides by 2.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Reversing the standard derivatives 3.5
Every differentiation result reads backwards as an integral: int e^x dx = e^x + C, int 1/x dx = ln|x| + C, int sin x dx = -cos x + C, int cos x dx = sin x + C, int sec^2 x dx = tan x + C.
A linear inside function divides by its derivative: int e^(kx) dx = e^(kx)/k + C, int sin(ax + b) dx = -cos(ax + b)/a + C, int (ax + b)^n dx = (ax + b)^(n+1)/(a(n+1)) + C for n not -1.
Worked example: the area under e^(2x) from 0 to 0.5 is [e^(2x)/2] = (e - 1)/2 = 0.8591 (rounded to 4 d.p.) - the chain multiplier 2 becomes a divisor.
int 1/(ax + b) dx = ln|ax + b|/a + C; the n = -1 case of the power pattern is always a logarithm, never a power.
Always check by differentiating your answer back - the derivative must return the integrand exactly, including every constant factor.
The f'(x)/f(x) logarithm pattern 3.5
When the numerator is the derivative of the denominator, int f'(x)/f(x) dx = ln|f(x)| + C - the reverse of the chain rule applied to ln f(x).
Worked example: int 2x/(x^2 + 1) dx = ln(x^2 + 1) + C, because the top is exactly the derivative of the bottom.
A constant multiple still works: int x/(x^2 + 1) dx = (1/2) ln(x^2 + 1) + C - supply the missing factor and compensate outside.
This pattern decides between a logarithm and a substitution: if adjusting a constant makes the top the derivative of the bottom, the answer is a logarithm.
Integration by substitution 3.5
Choose u as the inside function when its derivative sits alongside: int f(g(x)) g'(x) dx = int f(u) du. Replace every x, including dx, before integrating.
Worked example: int 2x e^(x^2) dx with u = x^2 becomes int e^u du = e^u, so the antiderivative is e^(x^2); from 0 to 1 the area is e - 1 = 1.7183 (rounded to 4 d.p.).
For definite integrals either convert the limits to u-values and never return to x, or substitute back and use the original limits - mixing the two is the classic mark-loser.
Trig example: int cos x sin^3 x dx with u = sin x becomes int u^3 du = sin^4 x/4 + C.
If the leftover x does not cancel after substitution, the chosen u is wrong - look for a different inside function.
Integration by parts 3.5
The product rule backwards: int u dv/dx dx = u v - int v du/dx dx. Choose u so its derivative is simpler - polynomials before exponentials and trig, ln x almost always as u.
Worked example: int x e^x dx with u = x gives x e^x - int e^x dx = (x - 1) e^x + C; from 0 to 1 the area is 0 - (-1) = 1 exactly.
Worked example: int x cos x dx = x sin x - int sin x dx = x sin x + cos x + C; from 0 to pi that is (0 - 1) - (0 + 1) = -2, a signed area.
int ln x dx is by parts in disguise with u = ln x and dv = 1: it gives x ln x - x + C.
Sometimes two rounds are needed (x^2 e^x), and sometimes the original integral reappears and is solved by rearranging (e^x sin x).
Trigonometric identities in integration 3.5
Even powers of sine and cosine have no direct antiderivative: rewrite first with cos^2 x = (1 + cos 2x)/2 or sin^2 x = (1 - cos 2x)/2.
Worked example: int cos^2 x dx = x/2 + sin(2x)/4 + C; from 0 to pi/2 the sine terms vanish, leaving pi/4 = 0.7854 (rounded to 4 d.p.).
int tan^2 x dx uses tan^2 x = sec^2 x - 1, giving tan x - x + C - identities first, integration second.
Products like sin 3x cos x rewrite with the product-to-sum identities into two sines, each integrated at sight.
Odd powers go by substitution instead: int sin^3 x dx = int (1 - cos^2 x) sin x dx with u = cos x.
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