Understand · explore · practise
Choose integration methods independently and combine them in original mixed practice. Check exact integrals, areas, numerical estimates, parameters and differential equations with optional hints and worked solutions.
Before you startThe preceding Pure 2 integration lessons, including numerical integration and differential equations.
01 / Choose a method before doing the algebra
Simplify → recognise → choose → calculate → verify
A product does not automatically need parts; a fraction does not automatically need partial fractions.
Try each question on paper before opening its hint or worked solution. Decide whether the question asks for an indefinite integral, a signed definite integral, a geometric area, a numerical estimate or a differential-equation solution. That decision affects both the method and the checks.
Find ∫x/(x² + 5) dx.
Choose a method, then check it. More than one route can be valid.
Explain your choice before looking at feedback. A recognised derivative and an explicit substitution can express the same idea. This activity checks a useful first step, not whether every possible longer route could work.
02 / Start with linearity and identities
Find ∫(3x² + 2e²ˣ − 4/x) dx, for x ≠ 0.
Integrate the three terms separately and retain a logarithmic absolute value.
x³ + e²ˣ − 4ln|x| + C, on an interval avoiding zero. Differentiation returns each original term.
Find ∫sin²(3x) dx.
Use sin²θ = (1 − cos 2θ)/2.
∫[1 − cos(6x)]/2 dx = x/2 − sin(6x)/12 + C. The factor 1/6 is needed when integrating cos(6x).
03 / Look for an inner derivative
Find ∫x/(x² + 5) dx.
The denominator has derivative 2x.
The answer is (1/2)ln(x² + 5) + C. Here x² + 5 is positive for all real x.
Find ∫xe^(x²+1) dx.
Compare x with the derivative of x² + 1.
(1/2)e^(x²+1) + C. Integration by parts would make this less direct.
04 / Change limits with the variable
Evaluate ∫₀¹ x(1 + x²)³ dx exactly.
Set u = 1 + x². The new limits are 1 and 2.
du = 2x dx, so the integral is (1/2)∫₁² u³ du = [u⁴/8]₁² = 15/8.
Evaluate ∫₀^(π/6) tan x dx.
Write tan x = sin x/cos x.
An antiderivative is −ln|cos x|. On these bounds cos x > 0, giving −ln(√3/2) = ln(2/√3).
05 / Choose the differentiated factor deliberately
Find ∫x cos(2x) dx.
Choose u = x and dv = cos(2x) dx.
v = sin(2x)/2. Hence ∫x cos(2x) dx = x sin(2x)/2 − ∫sin(2x)/2 dx = x sin(2x)/2 + cos(2x)/4 + C.
Evaluate ∫₁ᵉ ln x dx.
Use u = ln x and dv = dx.
[x ln x − x]₁ᵉ = 0 − (−1) = 1.
06 / Compare degrees, then inspect factors
Find ∫(3x + 5)/[(x + 1)(x + 2)] dx.
Write the integrand as A/(x + 1) + B/(x + 2).
3x + 5 = A(x + 2) + B(x + 1), giving A = 2, B = 1. The answer is 2ln|x + 1| + ln|x + 2| + C on an interval avoiding −1 and −2.
Find ∫x²/(x + 1) dx.
Divide x² by x + 1 first.
x²/(x + 1) = x − 1 + 1/(x + 1). Integrate to get x²/2 − x + ln|x + 1| + C, for an interval avoiding −1.
07 / Separate signed integral from geometric area
For y = x² − 4 on [−3, 3], find the signed integral and the total area between the curve and the x-axis.
Split at x = −2 and x = 2.
The signed integral is [x³/3 − 4x]₋₃³ = −6. The central area is ∫₋₂²(4 − x²) dx = 32/3. Each outer piece has area 7/3. Total area = 46/3 square units.
Find the finite area enclosed by y = x + 2 and y = x².
Find their intersections, then subtract the lower curve.
x² = x + 2 gives x = −1, 2. Between them the line is above the parabola. Area = ∫₋₁²(x + 2 − x²) dx = [x²/2 + 2x − x³/3]₋₁² = 9/2.
08 / Use numerical integration and judge the estimate
Use two equal strips to estimate ∫₀¹ eˣ dx. Is the estimate above or below the exact integral?
The ordinates are 1, e^(1/2), e and the width is 1/2.
T = (1/4)[1 + 2e^(1/2) + e] ≈ 1.75393. Since (eˣ)″ = eˣ > 0, the chords lie above the curve: T overestimates the exact value e − 1 ≈ 1.71828.
Apply two equal trapezia to ∫₋₁¹ x³ dx and compare with the exact integral.
The ordinates are −1, 0 and 1.
T = (1/2)[−1 + 2(0) + 1] = 0, equal to the exact integral by odd symmetry. The curve changes curvature, so this example does not prove that trapezia are always exact for cubics.
09 / Apply initial data and check the original equation
Solve y′ = 2xy with y(0) = 1.
Separate dy/y = 2x dx for the nonzero branch.
ln|y| = x² + C, giving y = Ae^(x²). The initial point gives A = 1. Thus y = e^(x²), valid for all real x. The equilibrium y = 0 exists but does not meet the initial condition.
Solve y′ = x/y with y(0) = 2.
Multiply by y, integrate and use the initial value.
y²/2 = x²/2 + C; the initial value gives y² = x² + 4. Choose y = √(x² + 4). It remains positive and solves the original equation for all real x.
10 / Treat an unknown bound or coefficient as part of the problem
Find a ≥ 0 such that ∫₀ᵃ(2x + 1) dx = 12.
Integrate first, then solve a quadratic.
a² + a = 12, so (a − 3)(a + 4) = 0. The restriction a ≥ 0 selects a = 3.
Find k such that ∫₀²(kx + 1) dx = 8.
Keep k constant during integration.
[kx²/2 + x]₀² = 2k + 2 = 8, so k = 3.
11 / Combine methods when the structure calls for it
Solve y′ = xe⁻ˣ/y with y(0) = 1.
Separate y dy = xe⁻ˣ dx, then integrate the right side by parts.
y²/2 = −(x + 1)e⁻ˣ + C. The initial point gives C = 3/2, so y = √[3 − 2(x + 1)e⁻ˣ]. The radicand has its minimum 1 at x = 0, so the positive branch exists for all real x. Differentiating verifies the original equation.
Evaluate ∫₀¹ x ln(1 + x²) dx exactly.
First use u = 1 + x²; then integrate ln u by parts.
The integral becomes (1/2)∫₁² ln u du = (1/2)[u ln u − u]₁² = ln 2 − 1/2. The new bounds and the factor 1/2 both matter.
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Section 1 of 11 · Choose a method before doing the algebra