# Integration (FP3)

> Edexcel International A-Level Further Mathematics · 2018 IAL FP3 (WFM03)
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp3-integration/

This guide covers all 6 core integration outcomes for Edexcel IAL FP3, including hyperbolic/inverse function integration, substitution, reduction formulae, arc length and surface area of revolution, with step-by-step worked examples and exam-specific tips.

**Prerequisites:** [Integration by parts and substitution (IAL A-level Mathematics Pure 3)](https://www.owlsprep.com/study/edexcel-ial-math-p3-integration/); [Differentiation of hyperbolic and inverse functions (FP3 Differentiation)](https://www.owlsprep.com/study/edexcel-ial-further-math-fp3-differentiation/)

## Learning objectives

- Integrate hyperbolic, inverse trigonometric and inverse hyperbolic functions using standard results and integration by parts
- Apply trigonometric/hyperbolic substitutions to evaluate integrals of standard reciprocal and surd forms
- Derive and apply reduction formulae for definite and indefinite integrals
- Calculate arc length and surface area of revolution for Cartesian and parametric curves

## Integrating Hyperbolic, Inverse Trigonometric and Hyperbolic Functions

Standard hyperbolic function integral results are provided in the formula book, so you can apply them directly to evaluate indefinite and definite integrals. For inverse trigonometric and inverse hyperbolic functions, use integration by parts with $u$ set equal to the inverse function, and $dv = dx$.

**Integration of inverse functions** — For any integrable inverse function $f^{-1}(x)$, use the integration by parts setup: $u = f^{-1}(x)$, $dv = dx$, so $du = \frac{1}{f'(f^{-1}(x))}dx$, $v = x$, giving $\int f^{-1}(x)dx = x f^{-1}(x) - \int \frac{x}{f'(f^{-1}(x))}dx$.

**Worked example:** Evaluate $\int_0^1 \text{arsinh } x dx$

1. Set $u = \text{arsinh }x$, $dv = dx$
2. $$du = \frac{1}{\sqrt{1+x^2}}dx, v = x$$
3. Apply integration by parts formula:
4. $$\int \text{arsinh }x dx = x \text{arsinh }x - \int \frac{x}{\sqrt{1+x^2}}dx$$
5. Evaluate the remaining integral using substitution $t = 1+x^2$, $dt = 2x dx$:
6. $$\int \frac{x}{\sqrt{1+x^2}}dx = \sqrt{1+x^2} + C$$
7. Apply limits 0 to 1:
8. $$[x \text{arsinh }x - \sqrt{1+x^2}]_0^1 = \text{arsinh }1 - \sqrt{2} + 1$$
9. Use standard result $\text{arsinh }1 = \ln(1 + \sqrt{2})$ to get the final value: $\ln(1+\sqrt{2}) - \sqrt{2} + 1 \approx 0.467$

> **Exam tip**
>
> You may use the standard surd integral results from the formula book directly for the remaining integral after applying integration by parts, to save time in exams.

*Calculator:* allowed

## Trigonometric and Hyperbolic Substitution for Integrals

Use standard substitutions to simplify integrals involving quadratic surds. All standard substitutions and corresponding integral results are given in the formula book. For more complex surd integrals, the required substitution will be provided in the question, so you do not need to invent new substitutions.

| Integrand Form | Substitution | Simplified Surd |
| --- | --- | --- |
| $1/\sqrt{a^2 + x^2}$ | $x = a \sinh u$ | $a \cosh u$ |
| $1/\sqrt{a^2 - x^2}$ | $x = a \sin \theta$ | $a \cos \theta$ |
| $1/\sqrt{x^2 - a^2}$ | $x = a \cosh u$ | $a \sinh u$ |
| $1/(a^2 + x^2)$ | $x = a \tan \theta$ | $a^2 \sec^2 \theta$ |

**Worked example:** Evaluate $\int \frac{1}{\sqrt{4x^2 - 9}}dx$ for $x > 3/2$

1. Rewrite the integrand to match the standard form:
2. $$\int \frac{1}{\sqrt{(2x)^2 - 3^2}}dx$$
3. Use substitution $2x = 3 \cosh u$, so $2dx = 3 \sinh u du$, $dx = \frac{3}{2}\sinh u du$
4. Substitute into the integral:
5. $$\int \frac{1}{\sqrt{9 \cosh^2 u - 9}} \times \frac{3}{2}\sinh u du = \int \frac{1}{3 \sinh u} \times \frac{3}{2}\sinh u du = \int \frac{1}{2} du$$
6. $$= \frac{1}{2}u + C = \frac{1}{2}\text{arcosh}\left(\frac{2x}{3}\right) + C$$
7. This matches the standard result from the formula book, so you could also write it directly as $\frac{1}{2}\ln\left(2x + \sqrt{4x^2 - 9}\right) + C$

*Calculator:* allowed

## Deriving and Applying Reduction Formulae

**Exam command terms**

Command terms for reduction formula questions have specific expectations:

- **Derive** — Show full working to get the recurrence relation from the definition of $I_n$, no marks are awarded for stating the relation without proof

- **Evaluate** — Apply the derived recurrence relation to base cases (usually $I_0$ or $I_1$) to get a numerical or exact value for the integral

**Reduction formula** — A recurrence relation that expresses $I_n$ (the integral of a function raised to power $n$) in terms of $I_{n-k}$ for $k < n$, usually $k=2$ for trigonometric and hyperbolic functions.

**Worked example:** Derive a reduction formula for $I_n = \int_0^{\pi/2} \sin^n x dx$, and use it to evaluate $I_4$

1. Split the integrand into two terms for integration by parts:
2. $$I_n = \int_0^{\pi/2} \sin^{n-1}x \times \sin x dx$$
3. Set $u = \sin^{n-1}x$, $dv = \sin x dx$, so $du = (n-1)\sin^{n-2}x \cos x dx$, $v = -\cos x$
4. Apply integration by parts:
5. $$I_n = [-\sin^{n-1}x \cos x]_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2}x \cos^2 x dx$$
6. The boundary term evaluates to 0, substitute $\cos^2 x = 1 - \sin^2 x$:
7. $$I_n = (n-1)\int_0^{\pi/2} \sin^{n-2}x (1 - \sin^2 x) dx = (n-1)I_{n-2} - (n-1)I_n$$
8. Rearrange to get the recurrence relation:
9. $$I_n + (n-1)I_n = (n-1)I_{n-2} \implies nI_n = (n-1)I_{n-2}$$
10. Evaluate $I_4$ using base case $I_0 = \int_0^{\pi/2} 1 dx = \pi/2$:
11. $$I_4 = \frac{3}{4}I_2 = \frac{3}{4} \times \frac{1}{2}I_0 = \frac{3}{8} \times \frac{\pi}{2} = \frac{3\pi}{16}$$

> **Exam tip**
>
> Always state the recurrence relation explicitly before applying it to base cases, as marks are awarded for the correct relation even if you make an error in the final evaluation.

*Calculator:* allowed

## Calculating Arc Length for Cartesian and Parametric Curves

Arc length formulae for Cartesian and parametric curves are given in the formula book. Polar form arc length is out of scope for FP3. Always simplify the expression inside the square root first, as it often simplifies to a perfect square.

**Worked example:** Calculate the arc length of the curve $y = \frac{2}{3}x^{3/2}$ between $x=0$ and $x=3$

1. Differentiate y with respect to x:
2. $$\frac{dy}{dx} = x^{1/2} = \sqrt{x}$$
3. Substitute into the Cartesian arc length formula:
4. $$s = \int_0^3 \sqrt{1 + (\frac{dy}{dx})^2} dx = \int_0^3 \sqrt{1 + x} dx$$
5. Evaluate the integral:
6. $$s = \left[\frac{2}{3}(1+x)^{3/2}\right]_0^3 = \frac{2}{3}(4^{3/2} - 1^{3/2}) = \frac{2}{3}(8 - 1) = \frac{14}{3} \approx 4.67$$

*Calculator:* allowed

## Calculating Surface Area of Revolution

Surface area of revolution about the x-axis is given by $S_x = 2\pi \int y ds$, where $ds$ is the arc length element matching the curve's form (Cartesian or parametric). Do not forget the $y$ term or the $ds$ term, as these are common sources of error.

**Worked example:** Find the surface area formed when the curve $y = \sqrt{x}$ for $0 \leq x \leq 4$ is rotated $2\pi$ radians about the x-axis

1. Differentiate y with respect to x:
2. $$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$$
3. Write the ds term for Cartesian form:
4. $$ds = \sqrt{1 + (\frac{dy}{dx})^2} dx = \sqrt{1 + \frac{1}{4x}} dx = \sqrt{\frac{4x + 1}{4x}} dx = \frac{\sqrt{4x + 1}}{2\sqrt{x}} dx$$
5. Substitute into the surface area formula:
6. $$S_x = 2\pi \int_0^4 y ds = 2\pi \int_0^4 \sqrt{x} \times \frac{\sqrt{4x + 1}}{2\sqrt{x}} dx$$
7. Simplify the integrand, the $\sqrt{x}$ terms cancel out:
8. $$S_x = \pi \int_0^4 \sqrt{4x + 1} dx$$
9. Evaluate the integral using substitution $u = 4x + 1$, $du = 4dx$:
10. $$S_x = \pi \times \frac{1}{4} \times \frac{2}{3}[(4x + 1)^{3/2}]_0^4 = \frac{\pi}{6}(17^{3/2} - 1) \approx 36.2$$

> **warning**
>
> Never use $dx$ in place of $ds$ in surface area calculations, as this will give you the wrong result. Always substitute the full $ds$ expression for the curve's form.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using $x = a \sinh u$ substitution for $\sqrt{x^2 - a^2}$ instead of $x = a \cosh u$
  - Why it fails: $\sinh^2 u + 1 = \cosh^2 u$, so $\sqrt{a^2 \sinh^2 u + a^2} = a \cosh u$, which works for $\sqrt{x^2 + a^2}$ not $\sqrt{x^2 - a^2}$
  - Correct: Use $x = a \cosh u$ for $\sqrt{x^2 - a^2}$, as $\cosh^2 u - 1 = \sinh^2 u$, simplifying the surd to $a \sinh u$ for $x \geq a$
- **Wrong:** Writing $S_x = 2\pi \int y dx$ instead of $2\pi \int y ds$ for surface area calculations
  - Why it fails: Surface area depends on the length of the curve segment, not just the x-axis interval, so omitting $ds$ gives an incorrect value
  - Correct: Always substitute the correct $ds$ form (Cartesian or parametric) matching the curve's equation before evaluating the integral
- **Wrong:** Splitting $I_n = \int \sin^n x dx$ as $u = \sin^n x$, $dv = dx$ when deriving reduction formulae
  - Why it fails: This split leads to an integral with $x \sin^{n-1}x \cos x$, which is more complex and does not produce a lower power of $\sin x$
  - Correct: Split powers of trigonometric/hyperbolic functions into a $(n-1)$ power term and a 1 power term for integration by parts to get a recurrence relation with $I_{n-2}$
- **Wrong:** Trying to integrate inverse functions directly instead of using integration by parts
  - Why it fails: Inverse functions have no elementary antiderivative when integrated directly, so this approach will not work
  - Correct: For integrals of inverse functions like $\int \arctan x dx$, set $u = $ inverse function, $dv = dx$, then evaluate the resulting simpler integral
- **Wrong:** Using incorrect limits for parametric arc length/surface area integrals
  - Why it fails: The parameter limits must correspond exactly to the start and end points of the curve segment being measured, not the x-axis limits
  - Correct: Check that you are using the correct parameter interval that maps to the curve endpoints given in the question

## Cheatsheet

| Concept | Formula / Technique | Exam Tip |
| --- | --- | --- |
| Integrate inverse functions | Use by parts: $u = f^{-1}(x)$, $dv = dx$ | Use formula book surd results directly for the resulting integral to save time |
| $\sqrt{a^2 + x^2}$ substitution | $x = a \sinh u$ | Simplifies to $a \cosh u$, result maps to $\text{arsinh}(x/a)$ |
| $\sqrt{a^2 - x^2}$ substitution | $x = a \sin \theta$ | Simplifies to $a \cos \theta$, result maps to $\arcsin(x/a)$ |
| $\sqrt{x^2 - a^2}$ substitution | $x = a \cosh u$ | Simplifies to $a \sinh u$, result maps to $\text{arcosh}(x/a)$ |
| Reduction formula derivation | Use by parts: $u = f(x)^{n-1}$, $dv = f(x)dx$ | State the recurrence relation explicitly before applying to base cases |
| Cartesian arc length | $s = \int \sqrt{1 + (dy/dx)^2} dx$ | Simplify the term inside the square root first, look for perfect squares |
| Parametric arc length | $s = \int \sqrt{(dx/dt)^2 + (dy/dt)^2} dt$ | Use parameter limits matching the curve endpoints, not x/y limits |
| Surface area (x-axis) | $S_x = 2\pi \int y ds$ | Use the $ds$ form matching the curve's equation (Cartesian/parametric) |

## What's next

Now that you have mastered FP3 integration techniques, you are ready to apply these skills to the remaining core FP3 topics, as well as synoptic questions across the Further Mathematics specification. Integration of hyperbolic functions and reduction formulae are often combined with differential equations topics in FP4, while arc length and surface area calculations may appear in synoptic questions linking to coordinate geometry in FP2. Be sure to practice full past paper questions for this topic, as exam questions typically combine multiple integration techniques in a single multi-part problem, worth 8-12 marks. You should also revise integration techniques from Pure 3 to ensure you can quickly recall substitution and by parts rules under timed exam conditions.

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