Integration (Pure 3) — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: Standard integrals for exponential, reciprocal and trigonometric functions, integration by substitution, integration by parts, partial fractions integration, and simplifying integrals using trigonometric identities.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trig).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Integration (Pure 3)?

Integration is the reverse process of differentiation, used to calculate areas under curves, volumes of revolution, solve differential equations, and evaluate quantities in applied math contexts. The P3 integration syllabus builds on the basic polynomial integration you learned in Pure 1 to cover more complex function types, with questions typically making up 15-20% of total Paper 3 marks. Indefinite integrals give a general antiderivative plus a constant of integration $+C$, while definite integrals evaluate to a numerical value between fixed upper and lower limits.

2. Standard integrals — $e^x,1/x,\sin ,\cos ,{\sec }^2$

All standard integrals are derived directly from differentiation rules you already know, so you can verify any rule by differentiating the result to get back the original integrand. The core P3 standard integrals are:

1. Exponential: $\int e^{kx}dx=\frac{1}{k}{e}^{kx}+C$, where $k$ is a non-zero constant. This comes from $\frac{d}{dx}{e}^{kx}=k{e}^{kx}$.
2. Reciprocal: $\int \frac{1}{x}dx=\ln |x|+C$. The absolute value is critical, as $\ln$ is only defined for positive inputs, so it covers both positive and negative values of $x$. Examiners regularly deduct marks for missing the absolute value.
3. Trigonometric:

• $\int \sin (kx)dx=-\frac{1}{k}\cos (kx)+C$ (note the negative sign, opposite to the derivative of $\cos$)
• $\int \cos (kx)dx=\frac{1}{k}\sin (kx)+C$
• $\int {\sec }^2(kx)dx=\frac{1}{k}\tan (kx)+C$

Worked Example

Find $\int (3e^{4x}+\frac{2}{x}-5{\sec }^2(2x))dx$: Integrate term by term: $$3\times \frac{1}{4}{e}^{4x}+2\ln |x|-5\times \frac{1}{2}\tan (2x)+C=\frac{3}{4}{e}^{4x}+2\ln |x|-\frac{5}{2}\tan (2x)+C$$

3. Integration by substitution

Integration by substitution is the reverse of the chain rule for differentiation, used for integrands that are the product of a composite function and its derivative. The steps are:

1. Choose $u$ as the inner function of the composite term (usually the expression inside brackets, exponents, or trig functions).
2. Differentiate $u$ with respect to $x$ to get $\frac{du}{dx}$, rearrange to solve for $dx=\frac{du}{f^{\prime }(x)}$.
3. Replace all $x$ terms in the integral with $u$ terms, simplifying to an integral you can solve with standard rules.
4. For indefinite integrals, substitute back $u=f(x)$ in the final result. For definite integrals, convert the upper and lower $x$ limits to $u$ limits immediately, so you do not need to convert back to $x$ at the end.

Worked Example

Find $\int 6x^2(x^3+2{)}^{5}dx$: Let $u=x^3+2$, so $\frac{du}{dx}=3x^2$ → $dx=\frac{du}{3x^2}$. Substitute into the integral: $$\int 6x^2{u}^5\times \frac{du}{3x^2}=\int 2u^{5}du=2\times \frac{u^6}{6}+C=\frac{(x^3+2{)}^6}{3}+C$$

Exam tip: For harder questions, examiners will give you the required substitution, so follow the given $u$ exactly to avoid mistakes.

4. Integration by parts — $\int udv=uv-\int vdu$

Integration by parts is the reverse of the product rule for differentiation, used for integrands that are the product of two unrelated function types. The formula is derived by rearranging the product rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, then integrating both sides.

To choose which term is $u$ (the term you differentiate) and which is $dv$ (the term you integrate), use the LIATE priority rule:

1. Logarithmic functions
2. Inverse trigonometric functions
3. Algebraic functions
4. Trigonometric functions
5. Exponential functions The term that appears higher in the list is your $u$, the rest is $dv$.

Worked Example

Find $\int x{e}^{3x}dx$: Algebraic $x$ comes before exponential $e^{3x}$ in LIATE, so $u=x$, $dv=e^{3x}dx$. Differentiate $u$: $du=dx$. Integrate $dv$: $v=\frac{1}{3}{e}^{3x}$. Apply the formula: $$uv-\int vdu=\frac{1}{3}x{e}^{3x}-\int \frac{1}{3}{e}^{3x}dx=\frac{1}{3}x{e}^{3x}-\frac{1}{9}{e}^{3x}+C$$

5. Integrating using partial fractions

Partial fractions integration is used for rational functions (ratios of two polynomials) that cannot be integrated directly. The steps are:

1. Check if the fraction is proper: the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first to write the expression as a polynomial plus a proper rational function.
2. Factor the denominator into linear or irreducible quadratic factors.
3. Decompose the proper fraction into partial fractions, each with a constant numerator and linear denominator.
4. Integrate each partial fraction separately using the standard $\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+C$ rule.

Worked Example

Find $\int \frac{5x+1}{x^2+x-6}dx$: First factor the denominator: $x^2+x-6=(x+3)(x-2)$. Decompose into partial fractions: $$\frac{5x+1}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2}$$ Multiply both sides by the denominator: $5x+1=A(x-2)+B(x+3)$. Substitute $x=-3$: $-15+1=A(-5)$ → $A=\frac{14}{5}$. Substitute $x=2$: $10+1=B(5)$ → $B=\frac{11}{5}$. Integrate: $$\int \left(\frac{14/5}{x+3}+\frac{11/5}{x-2}\right)dx=\frac{14}{5}\ln |x+3|+\frac{11}{5}\ln |x-2|+C$$ You can combine into a single logarithm for simplified answers: $\ln |(x+3{)}^{14}(x-2{)}^{11}{|}^{1/5}+C$.

6. Trigonometric identities to simplify integrals

Many trigonometric integrands cannot be solved with standard rules directly, so you use identities to rewrite them into integrable forms. The most commonly tested identities in P3 are:

1. Double angle identities: Used to integrate ${\sin }^2(kx)$ and ${\cos }^2(kx)$:

• $\cos (2kx)=2{\cos }^2(kx)-1$ → ${\cos }^2(kx)=\frac{1+\cos (2kx)}{2}$
• $\cos (2kx)=1-2{\sin }^2(kx)$ → ${\sin }^2(kx)=\frac{1-\cos (2kx)}{2}$

2. Pythagorean identities: ${\sin }^{2}x+{\cos }^{2}x=1$, ${\sec }^{2}x=1+{\tan }^{2}x$, used to reduce higher powers of trig functions.
3. Product-to-sum identities: Used for products of $\sin$ and $\cos$ terms, e.g. $\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)]$.

Worked Example

Find $\int {\cos }^2(4x)dx$: Use the double angle identity: ${\cos }^2(4x)=\frac{1+\cos (8x)}{2}$. Integrate: $$\int \left(\frac{1}{2}+\frac{1}{2}\cos (8x)\right)dx=\frac{x}{2}+\frac{1}{16}\sin (8x)+C$$

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Forgetting the constant of integration $+C$ for indefinite integrals. Students rush through final steps and omit it, losing 1 mark per question. Correct move: Add $+C$ immediately after finishing integration, before simplifying your answer.
• Pitfall 2: Mixing up signs when integrating $\sin$ and $\cos$. Students confuse integration rules with differentiation rules. Correct move: Write down the derivative rule next to the integral rule when practicing, e.g. $\frac{d}{dx}\cos x=-\sin x$ so $\int \sin xdx=-\cos x+C$.
• Pitfall 3: Forgetting to adjust limits for definite integral substitution. Students substitute $u$ into the integrand but keep the original $x$ limits, leading to wrong numerical answers. Correct move: Convert upper and lower limits to $u$ values as soon as you define $u$, before evaluating the integral.
• Pitfall 4: Trying to decompose improper fractions into partial fractions without long division first. Students skip the proper fraction check, leading to incorrect partial fraction values. Correct move: Always compare the degree of numerator and denominator first; divide if the numerator degree is equal or higher.
• Pitfall 5: Omitting the absolute value in $\ln |x|$ terms. Students write $\ln x$ instead of $\ln |x|$, which is only defined for positive $x$. Correct move: Add absolute value bars around the argument of every logarithm from integration, unless the question specifies the domain is positive.

8. Practice Questions (Paper 3 Style)

Question 1 (5 marks)

a) Find the indefinite integral $\int (2e^{5x}+\frac{4}{x}+3{\sec }^2(2x))dx$. (3 marks) b) Evaluate the definite integral ${\int }_1^4(3x+\frac{2}{x})dx$, giving your answer in exact form. (2 marks)

Solution

a) Integrate term by term using standard rules: $$\int 2e^{5x}dx=2\times \frac{1}{5}{e}^{5x}=\frac{2}{5}{e}^{5x}$$ $$\int \frac{4}{x}dx=4\ln |x|$$ $$\int 3{\sec }^2(2x)dx=3\times \frac{1}{2}\tan (2x)=\frac{3}{2}\tan (2x)$$ Final answer: $\frac{2}{5}{e}^{5x}+4\ln |x|+\frac{3}{2}\tan (2x)+C$ (3 marks, 1 per term, 1 for $+C$) b) Integrate first: $\int (3x+\frac{2}{x})dx=\frac{3}{2}{x}^2+2\ln |x|$ Evaluate at limits: $(\frac{3}{2}(16)+2\ln 4)-(\frac{3}{2}(1)+2\ln 1)=24+2\ln 4-1.5=\frac{45}{2}+4\ln 2$ (2 marks)

Question 2 (6 marks)

a) Use integration by parts to find $\int x\ln xdx$. (4 marks) b) Evaluate ${\int }_0^{\pi /2}x\sin xdx$, giving your answer exact. (2 marks)

Solution

a) Logarithmic $\ln x$ comes before algebraic $x$ in LIATE, so $u=\ln x$, $dv=xdx$. $du=\frac{1}{x}dx$, $v=\frac{x^2}{2}$. Apply the formula: $$uv-\int vdu=\frac{x^2\ln x}{2}-\int \frac{x^2}{2}\times \frac{1}{x}dx=\frac{x^2\ln x}{2}-\int \frac{x}{2}dx$$ Integrate remaining term: $\frac{x^2\ln x}{2}-\frac{x^2}{4}+C$ (4 marks) b) $u=x$, $dv=\sin xdx$, $du=dx$, $v=-\cos x$: $$-x\cos x+\int \cos xdx=-x\cos x+\sin x$$ Evaluate at limits: $(-\frac{\pi }{2}\cos (\frac{\pi }{2})+\sin (\frac{\pi }{2}))-(0+\sin 0)=0+1-0=1$ (2 marks)

Question 3 (7 marks)

a) Express $\frac{8x-9}{x^2-3x-4}$ as partial fractions. (3 marks) b) Hence find $\int \frac{8x-9}{x^2-3x-4}dx$, giving your answer as a single logarithm. (4 marks)

Solution

a) Factor denominator: $x^2-3x-4=(x-4)(x+1)$ $$\frac{8x-9}{(x-4)(x+1)}=\frac{A}{x-4}+\frac{B}{x+1}$$ $8x-9=A(x+1)+B(x-4)$. Substitute $x=4$: $32-9=5A$ → $A=\frac{23}{5}$. Substitute $x=-1$: $-8-9=-5B$ → $B=\frac{17}{5}$. Partial fractions: $\frac{23/5}{x-4}+\frac{17/5}{x+1}$ (3 marks) b) Integrate: $$\frac{23}{5}\ln |x-4|+\frac{17}{5}\ln |x+1|+C=\frac{1}{5}\ln |(x-4{)}^{23}(x+1{)}^{17}|+C$$ (4 marks)

9. Quick Reference Cheatsheet

10. What's Next

Mastering these integration techniques is non-negotiable for success in the rest of the A-Level Mathematics Paper 3 syllabus. You will use these rules to solve first-order differential equations, calculate volumes of revolution around both the x and y axes, and evaluate complex integrals in applied topics. If you are taking Mechanics (Paper 4) or Statistics (Paper 5), integration is also used to solve kinematics problems and calculate probabilities from probability density functions respectively, so a strong foundation here will save you significant time later in your course.

If you struggle with any of the rules, need more custom practice questions tailored to your weak points, or want to test your knowledge with full mock exam sections, you can ask Ollie for help anytime on the homepage. We also have dedicated study guides for all other A-Level Mathematics Paper 3 topics, from complex numbers to vectors, to help you get full marks on your exam.

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