Introduction to A2 Pure Integration
Welcome to one of the most rewarding and vital chapters in CCEA A2 1: Pure Mathematics. In AS Mathematics, you learned that integration represents the reverse of differentiation and allows us to find the area under a curve. In A2, we expand your mathematical toolkit dramatically.
You will learn how to integrate complex products, quotients, and trigonometric functions, calculate 3D Volumes of Revolution, and solve real-world rate-of-change problems using Differential Equations.
Don't worry if these new techniques look intimidating at first. By breaking each method down into clear, structured steps, you will quickly learn to recognize which technique to use for any given exam question.
A Quick Note on Notation: In CCEA examinations, the natural logarithm is written strictly as \(\ln x\) (which represents \(\log_e x\)). Always use this notation in your solutions.
---1. Standard Integrals and the Reverse Chain Rule
Before tackling advanced techniques, you must be completely fluent with the standard integrals provided in your CCEA Formulae and Tables booklet, as well as those you are expected to know automatically.
A. Standard Integrals Checklist
• Exponential functions: \(\int e^{kx} \, dx = \frac{1}{k}e^{kx} + c\)
• Reciprocal function: \(\int \frac{1}{x} \, dx = \ln|x| + c\)
• Sine: \(\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx) + c\)
• Cosine: \(\int \cos(kx) \, dx = \frac{1}{k}\sin(kx) + c\)
• Secant squared: \(\int \sec^2(kx) \, dx = \frac{1}{k}\tan(kx) + c\)
• Cosecant cotangent: \(\int \operatorname{cosec}(kx)\cot(kx) \, dx = -\frac{1}{k}\operatorname{cosec}(kx) + c\)
• Secant tangent: \(\int \sec(kx)\tan(kx) \, dx = \frac{1}{k}\sec(kx) + c\)
• Cosecant squared: \(\int \operatorname{cosec}^2(kx) \, dx = -\frac{1}{k}\cot(kx) + c\)
B. The Reverse Chain Rule (Integration by Inspection)
When an integrand consists of a composite function multiplied by something closely related to the derivative of the inside function, you can integrate directly by inspection.
Form 1: Power of a Function
\(\int f'(x)[f(x)]^n \, dx = \frac{[f(x)]^{n+1}}{n+1} + c \quad (n \neq -1)\)
Example: Evaluate \(\int 2x(x^2 + 5)^4 \, dx\).
Notice that the derivative of the inside function \((x^2 + 5)\) is precisely \(2x\). Therefore:
\(\int 2x(x^2 + 5)^4 \, dx = \frac{(x^2 + 5)^5}{5} + c\)
Form 2: Logarithmic Form (Derivative over Function)
\(\int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + c\)
Example: Evaluate \(\int \frac{3x^2}{x^3 + 7} \, dx\).
The numerator is exactly the derivative of the denominator. Therefore:
\(\int \frac{3x^2}{x^3 + 7} \, dx = \ln|x^3 + 7| + c\)
Key Takeaway: Whenever you see a fraction, always check if the numerator is a scalar multiple of the denominator's derivative. If it is, the integral evaluates directly to a natural logarithm.
---2. Trigonometric Integration Techniques
Examiners frequently test your ability to use trigonometric identities to rewrite expressions that cannot be integrated directly.
Integrating Even Powers of Sine and Cosine
Crucial Examiner Warning: A very common mistake is treating \(\int \sin^2 x \, dx\) as \(\frac{\sin^3 x}{3}\). This is mathematically incorrect because the chain rule does not work this way in reverse without an accompanying \(\cos x\) term!
To integrate \(\sin^2 x\) or \(\cos^2 x\), you must rewrite them using the double-angle identities:
• \(\cos(2x) = 1 - 2\sin^2 x \implies \sin^2 x = \frac{1 - \cos(2x)}{2}\)
• \(\cos(2x) = 2\cos^2 x - 1 \implies \cos^2 x = \frac{1 + \cos(2x)}{2}\)
Worked Example: Find \(\int \sin^2(3x) \, dx\).
Step 1: Use the identity \(\sin^2(3x) = \frac{1 - \cos(6x)}{2} = \frac{1}{2} - \frac{1}{2}\cos(6x)\).
Step 2: Integrate term by term:
\(\int \left( \frac{1}{2} - \frac{1}{2}\cos(6x) \right) dx = \frac{1}{2}x - \frac{1}{12}\sin(6x) + c\)
Key Takeaway: For \(\sin^2(kx)\) and \(\cos^2(kx)\), immediately substitute the double-angle cosine identity before attempting to integrate.
---3. Integration by Substitution
Integration by substitution is the integral equivalent of the chain rule in differentiation. It transforms an awkward integral in terms of \(x\) into a simpler integral in terms of a new variable, \(u\).
A. The General Method
1. Identify the substitution \(u = g(x)\) (either given in the exam question or chosen by inspection).
2. Differentiate \(u\) to find \(\frac{du}{dx}\), and rearrange to express \(dx\) in terms of \(du\).
3. Substitute all expressions of \(x\) so that the entire integral is in terms of \(u\).
4. Integrate with respect to \(u\).
5. If indefinite, convert back to \(x\). If definite, evaluate using the new \(u\)-limits.
B. Definite Integrals: Always Transform Your Limits!
When evaluating \(\int_a^b f(x) \, dx\) using \(u = g(x)\), convert your limits immediately:
• Lower limit: when \(x = a \implies u = g(a)\)
• Upper limit: when \(x = b \implies u = g(b)\)
\(\int_a^b f(g(x))g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du\)
Worked Example: Evaluate \(\int_0^1 x(x^2 + 1)^3 \, dx\) using the substitution \(u = x^2 + 1\).
Step 1: Differentiate the substitution: \(\frac{du}{dx} = 2x \implies dx = \frac{du}{2x}\).
Step 2: Change the limits:
• When \(x = 0 \implies u = (0)^2 + 1 = 1\)
• When \(x = 1 \implies u = (1)^2 + 1 = 2\)
Step 3: Substitute everything into the integral:
\(\int_1^2 x \cdot u^3 \cdot \frac{du}{2x} = \frac{1}{2} \int_1^2 u^3 \, du\)
Step 4: Integrate and evaluate directly with the \(u\)-limits:
\(\frac{1}{2} \left[ \frac{u^4}{4} \right]_1^2 = \frac{1}{8} \left[ 2^4 - 1^4 \right] = \frac{1}{8} [16 - 1] = \frac{15}{8}\)
Key Takeaway: Always change the limits of integration at the very start when using substitution on definite integrals. This avoids having to substitute back to \(x\) at the end.
---4. Integration by Parts
Integration by parts is used to integrate products of two functions where substitution does not apply.
The Formula
\(\int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx\)
Choosing Which Term is \(u\)
To choose which function should be \(u\), use the priority order LATE:
• L - Logarithmic functions (e.g., \(\ln x\)) — Highest priority for \(u\)
• A - Algebraic functions (e.g., \(x\), \(x^2\))
• T - Trigonometric functions (e.g., \(\sin x\), \(\cos x\))
• E - Exponential functions (e.g., \(e^x\)) — Lowest priority for \(u\) (best for \(\frac{dv}{dx}\))
Special Case 1: Integrating \(\ln x\)
To integrate \(\int \ln x \, dx\), we treat the integrand as the product \(1 \cdot \ln x\):
Let \(u = \ln x \implies \frac{du}{dx} = \frac{1}{x}\)
Let \(\frac{dv}{dx} = 1 \implies v = x\)
Applying the formula:
\(\int \ln x \, dx = x\ln x - \int x \cdot \frac{1}{x} \, dx = x\ln x - \int 1 \, dx = x\ln x - x + c\)
Special Case 2: Repeated Integration by Parts
For integrals like \(\int x^2 e^x \, dx\), you will need to apply the parts formula twice to reduce the power of \(x\) from \(x^2\) to \(x\), and then to a constant.
Special Case 3: Cyclic Integrals
For integrals like \(\int e^x \sin x \, dx\), applying parts twice returns a multiple of the original integral. You can then collect the original integral algebraically on one side of the equation to solve it.
Common Examiner Pitfall: Watch your minus signs! When calculating \(uv - \int v \frac{du}{dx} \, dx\), if \(v\) or \(\frac{du}{dx}\) contains a negative sign, ensure you don't lose track of double negatives.
Key Takeaway: Integration by parts undoes the product rule. Always differentiate \(u\) down to a simpler form and integrate \(\frac{dv}{dx}\).
---5. Integration Using Partial Fractions
When integrating rational functions (algebraic fractions) where the denominator factorises, we split the fraction into Partial Fractions first.
A. Proper vs. Improper Rational Functions
• Proper Fraction: Degree of numerator \(<\) degree of denominator. Split directly into partial fractions.
• Improper Fraction: Degree of numerator \(\ge\) degree of denominator. You must carry out polynomial long division first before splitting the remainder into partial fractions.
B. Decomposition Types
1. Linear Factors:
\(\frac{px + q}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}\)
Each term integrates to a natural logarithm: \(\int \frac{A}{x-a} \, dx = A\ln|x-a| + c\)
2. Repeated Linear Factors:
\(\frac{px^2 + qx + r}{(x-a)(x-b)^2} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{(x-b)^2}\)
Remember how to integrate the repeated factor: \(\int \frac{C}{(x-b)^2} \, dx = \int C(x-b)^{-2} \, dx = -\frac{C}{x-b} + c\)
Key Takeaway: Always check the degree of the numerator and denominator first. Linear denominators integrate to \(\ln\), but repeated linear terms integrate using the standard power rule.
---6. Geometric Applications: Areas and Volumes of Revolution
A. Area Between Two Curves
To find the area bounded between two curves \(y_1(x)\) and \(y_2(x)\) from \(x = a\) to \(x = b\):
\(\text{Area} = \int_a^b |y_1(x) - y_2(x)| \, dx\)
In practice, subtract the 'lower' curve from the 'upper' curve over the interval: \(\int_a^b (y_{\text{upper}} - y_{\text{lower}}) \, dx\).
B. Volumes of Revolution
When a 2D region is rotated through \(2\pi\) radians (\(360^\circ\)) around an axis, it sweeps out a 3D solid of revolution.
1. Rotation About the \(x\)-axis
\(V = \pi \int_a^b y^2 \, dx\)
2. Rotation About the \(y\)-axis
\(V = \pi \int_c^d x^2 \, dy\)
Note: For rotation about the \(y\)-axis, rearrange the equation to make \(x^2\) the subject in terms of \(y\), and use the \(y\)-coordinate limits \(c\) and \(d\).
3. Volume Between Two Curves (Hollow Solids)
When the region between two curves \(y_1\) and \(y_2\) (where \(y_1 \ge y_2\)) is rotated about the \(x\)-axis:
\(V = \pi \int_a^b (y_1^2 - y_2^2) \, dx\)
Crucial Examiner Warning: A major error in exam papers is writing \(\pi \int (y_1 - y_2)^2 \, dx\). You must square each function individually before subtracting: \(\pi \int (y_1^2 - y_2^2) \, dx\). Also, do not forget to multiply by \(\pi\)!
Key Takeaway: For volumes of revolution, think "pi times the integral of (radius squared)". Always square the function before integrating.
---7. First Order Differential Equations
A differential equation is an equation containing derivatives. In A2 1, you solve first-order differential equations using the method of Separation of Variables.
A. Separation of Variables
If \(\frac{dy}{dx} = g(x)h(y)\), we separate all \(y\) terms (including \(dy\)) to one side, and all \(x\) terms (including \(dx\)) to the other side:
\(\int \frac{1}{h(y)} \, dy = \int g(x) \, dx\)
B. General Solution vs. Particular Solution
• General Solution: The result of integrating both sides, containing an arbitrary constant of integration, \(+ c\).
• Particular Solution: Found by substituting given initial conditions or boundary conditions (e.g., \(y = y_0\) when \(x = x_0\)) to calculate the specific numerical value of \(c\).
C. Real-World Contexts: Exponential Growth and Decay
Rate models often state that the rate of change of a population or quantity \(P\) is proportional to the quantity itself:
\(\frac{dP}{dt} = kP \implies \int \frac{1}{P} \, dP = \int k \, dt \implies \ln|P| = kt + c \implies P = Ae^{kt}\) (where \(A = e^c\)).
Worked Example: Solve the differential equation \(\frac{dy}{dx} = 2xy^2\), given that \(y = 1\) when \(x = 0\).
Step 1: Separate variables:
\(\frac{1}{y^2} \, dy = 2x \, dx \implies y^{-2} \, dy = 2x \, dx\)
Step 2: Integrate both sides:
\(\int y^{-2} \, dy = \int 2x \, dx \implies -\frac{1}{y} = x^2 + c\)
Step 3: Apply the initial condition (\(x = 0, y = 1\)) to find \(c\):
-\frac{1}{1} = 0^2 + c \implies c = -1
Step 4: State the particular solution explicitly for \(y\):
-\frac{1}{y} = x^2 - 1 \implies \frac{1}{y} = 1 - x^2 \implies y = \frac{1}{1 - x^2}
Examiner Warning: Add the constant \(+ c\) immediately after integrating. Never add \(+ c\) at the end after rearranging for \(y\), as this will give an incorrect equation.
Key Takeaway: Separate variables cleanly so that \(dy\) and \(dx\) are always in the numerators. Integrate both sides, add \(+ c\) straight away, and use boundary conditions to find the particular solution.
---8. Summary: Choosing the Right Integration Method
When faced with an integral in your CCEA A2 1 exam, run through this quick decision checklist:
1. Is it a standard form from the Formulae Booklet? (e.g., \(e^{kx}\), \(\sec^2(kx)\), \(\frac{1}{x}\)) \(\implies\) Integrate directly.
2. Is it \(\sin^2(kx)\) or \(\cos^2(kx)\)? \(\implies\) Use double-angle identities.
3. Is the numerator the derivative of the denominator? \(\implies\) Use \(\ln|f(x)|\).
4. Is it a product where one part differentiates away? \(\implies\) Use Integration by Parts (LATE rule).
5. Is it an algebraic fraction with factorisable denominator? \(\implies\) Use Partial Fractions (check if long division is needed first).
6. Is there a composite bracket or specific substitution given? \(\implies\) Use Integration by Substitution (and transform your limits!).