Integration

Revise Integration for Pure Mathematics 2 P2 (AS Level) — revision notes and instant AI marking.

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Cambridge International AS & A Level Mathematics 9709 — Pure Mathematics 2

Integration

The big idea: Reverse every new derivative to get new integrals. When a function has no direct reverse, such as sin²x, an identity turns it into one that does; when nothing works, the trapezium rule gives an estimate.

Summary — What This Chapter Covers

∫e^ax+b dx = (1/a)e^ax+b + c and ∫1/(ax + b) dx = (1/a) ln|ax + b| + c.
∫sin(ax + b) dx = −(1/a) cos(ax + b) + c and ∫cos(ax + b) dx = (1/a) sin(ax + b) + c.
∫sec²(ax + b) dx = (1/a) tan(ax + b) + c.
Use cos 2x identities for sin²x and cos²x; use sec²x − 1 for tan²x.
Trapezium rule: area ≈ ½h[y₀ + y_n + 2(y₁ + … + y_n−1)].
It over-estimates when the curve bends upwards (convex) and under-estimates when it bends downwards (concave).

1. Integrating Exponential and Trigonometric Functions

Standard integrals

Reverse the new derivatives

Each standard integral comes from a derivative in the previous chapter; divide by the coefficient of x inside. The general method of substitution is not needed in Paper 2. For sin²x and cos²x, rewrite using the double angle formulae before integrating.

Standard integrals ∫e^ax+b dx = (1/a)e^ax+b · ∫1/(ax + b) dx = (1/a) ln|ax + b|
∫sin(ax + b) dx = −(1/a) cos(ax + b) · ∫cos(ax + b) dx = (1/a) sin(ax + b) · ∫sec²(ax + b) dx = (1/a) tan(ax + b)

Identities for integration sin²x = ½(1 − cos 2x) · cos²x = ½(1 + cos 2x) · tan²x = sec²x − 1

Worked Example

Find the exact values of (a) ∫₁³ 1/(2x − 1) dx, (b) ∫₀^π/2 sin²x dx.

Practice Question 1

Find ∫cos²(2x) dx.

Practice Question 2

Find the exact values of ∫₀¹ e^2x dx and ∫₀^π/8 sec²2x dx.

Practice Question 3

Find the exact value of ∫₀^π/4 tan²x dx.

2. The Trapezium Rule

Numerical integration

Strips of trapezia under the curve

Split [a, b] into n strips of equal width h = (b − a)/n. Note that n is the number of strips, so there are n + 1 ordinates. Work to more decimal places than the answer needs and round at the end.

On a sketch, the straight tops of the trapezia lie above a curve that bends upwards, so the rule over-estimates; they lie below a curve that bends downwards, so it under-estimates. More strips give a better estimate.

Trapezium rule ∫_a^b y dx ≈ ½h[y₀ + y_n + 2(y₁ + y₂ + … + y_n−1)], h = (b − a)/n

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Integration

Integration

Reverse the new derivatives

Strips of trapezia under the curve

More Pure Mathematics 2 topics