Further Calculus PM2

📚 Further Calculus - Complete Guide

Pure Mathematics 3| Cambridge AS & A Level

8.1 Derivative of tan⁻¹(x)

Understanding tan⁻¹(x):
The inverse tangent function (arctan) reverses the tangent operation.
If tan(θ) = x, then θ = tan⁻¹(x)

Derivation

Let y = tan⁻¹(x)

Then: tan(y) = x

Differentiate both sides:

sec²(y) × dy/dx = 1

Therefore: dy/dx = 1/sec²(y)

Using identity: sec²(y) = 1 + tan²(y) = 1 + x²

Result: dy/dx = 1/(1 + x²)

KEY FORMULA:

d/dx[tan⁻¹(x)] = 1/(1 + x²)

With chain rule:
d/dx[tan⁻¹(f(x))] = f'(x)/(1 + [f(x)]²)

Example 1: Differentiate tan⁻¹(3x)

Solution:
Let f(x) = 3x, so f'(x) = 3
d/dx[tan⁻¹(3x)] = 3/(1 + 9x²)

Example 2: Differentiate tan⁻¹(√x)

Solution:
f(x) = x^(1/2), f'(x) = 1/(2√x)
d/dx[tan⁻¹(√x)] = [1/(2√x)] / [1 + x]
= 1/[2√x(1 + x)]

8.2 Integration of 1/(x² + a²)

The Reverse Process:
Since d/dx[tan⁻¹(x)] = 1/(1 + x²), we can integrate backwards!

KEY FORMULAS:

∫ 1/(1 + x²) dx = tan⁻¹(x) + c

∫ 1/(x² + a²) dx = (1/a)tan⁻¹(x/a) + c

Why Does This Work?

Verify by differentiating:

d/dx[(1/a)tan⁻¹(x/a)] = (1/a) × 1/(1 + x²/a²) × (1/a)

                       = 1/(a² + x²) ✓

Example 1: Find ∫ 1/(x² + 25) dx

Solution:
Here a² = 25, so a = 5
∫ 1/(x² + 25) dx = (1/5)tan⁻¹(x/5) + c

Example 2: Evaluate ∫₀² 1/(x² + 4) dx

Solution:
a² = 4, so a = 2
= [(1/2)tan⁻¹(x/2)]₀²
= (1/2)tan⁻¹(1) - 0
= (1/2)(π/4) = π/8

Example 3: Find ∫ 1/(x² + 2x + 3) dx

Solution:
Complete the square: x² + 2x + 3 = (x + 1)² + 2
Let u = x + 1, then:
= ∫ 1/(u² + 2) du
= (1/√2)tan⁻¹(u/√2) + c
= (1/√2)tan⁻¹[(x + 1)/√2] + c

8.3 Integration of kf'(x)/f(x)

The Pattern:
Remember: d/dx[ln(f(x))] = f'(x)/f(x)
So integrating reverses this process!

KEY FORMULA:

∫ f'(x)/f(x) dx = ln|f(x)| + c

∫ kf'(x)/f(x) dx = k ln|f(x)| + c

How to Recognize This Pattern

Look for:

• A fraction (numerator/denominator)
• The numerator is (or can be adjusted to be) the derivative of the denominator
• The denominator is a function, not just x

Example 1: Find ∫ 2x/(x² + 3) dx

Solution:
Let f(x) = x² + 3
Then f'(x) = 2x ✓ (matches numerator!)
∫ 2x/(x² + 3) dx = ln|x² + 3| + c
= ln(x² + 3) + c (always positive)

Example 2: Find ∫ tan(x) dx

Solution:
tan(x) = sin(x)/cos(x)
Let f(x) = cos(x), f'(x) = -sin(x)
∫ tan(x) dx = ∫ -(-sin(x))/cos(x) dx
= -ln|cos(x)| + c

Example 3: Find ∫ x/(x² + 1) dx

Solution:
f(x) = x² + 1, f'(x) = 2x
Numerator is x, need 2x. Multiply by 1/2:
∫ x/(x² + 1) dx = (1/2)∫ 2x/(x² + 1) dx
= (1/2)ln(x² + 1) + c

8.4 Integration by Substitution

The Big Idea:
Transform a difficult integral into an easier one by changing variables.
Think of it like straightening a curved road to measure it!

The Method

Steps for Substitution:

1. Choose substitution u = g(x)
2. Find du/dx, rearrange to get dx
3. Replace ALL x terms with u terms
4. Integrate with respect to u
5. Substitute back to get answer in x

For definite integrals: Change the limits too!

Example 1: Find ∫ 4x(2x + 1)^(1/2) dx using u = 2x + 1

Solution:
u = 2x + 1
du/dx = 2, so dx = du/2
x = (u - 1)/2

∫ 4x√(2x + 1) dx = ∫ 4(u-1)/2 × √u × du/2
= ∫ (u - 1)√u du
= ∫ (u^(3/2) - u^(1/2)) du
= (2/5)u^(5/2) - (2/3)u^(3/2) + c
= (2/5)(2x+1)^(5/2) - (2/3)(2x+1)^(3/2) + c

Example 2: Find ∫ cos(x)sin⁵(x) dx

Solution:
Let u = sin(x)
du/dx = cos(x), so dx = du/cos(x)

∫ cos(x)sin⁵(x) dx = ∫ cos(x) × u⁵ × du/cos(x)
= ∫ u⁵ du
= (1/6)u⁶ + c
= (1/6)sin⁶(x) + c

Example 3 (Definite): Evaluate ∫₀³ x√(x + 1) dx using u = x + 1

Solution:
u = x + 1, so x = u - 1, dx = du
When x = 0: u = 1
When x = 3: u = 4

∫₀³ x√(x + 1) dx = ∫₁⁴ (u - 1)√u du
= ∫₁⁴ (u^(3/2) - u^(1/2)) du
= [(2/5)u^(5/2) - (2/3)u^(3/2)]₁⁴
= [64/5 - 16/3] - [2/5 - 2/3]
= 116/15

Example 4 (Trig Substitution): Find ∫ 8/(x² + 4) dx using x = 2tan(θ)

Solution:
x = 2tan(θ)
dx = 2sec²(θ) dθ
x² + 4 = 4tan²(θ) + 4 = 4sec²(θ)

∫ 8/(x² + 4) dx = ∫ 8/(4sec²(θ)) × 2sec²(θ) dθ
= ∫ 4 dθ = 4θ + c
= 4tan⁻¹(x/2) + c

8.5 Partial Fractions in Integration

The Strategy:
Break complicated fractions into simpler fractions that are easy to integrate.

Types of Partial Fractions

Type	Form
Linear factors	A/(x + a) + B/(x + b)
Repeated linear	A/(x + a) + B/(x + a)²
Irreducible quadratic	A/(x + a) + (Bx + C)/(x² + b)

Example 1: Find ∫ 3/[(x - 1)(x + 3)] dx

Solution:
3/[(x-1)(x+3)] = A/(x-1) + B/(x+3)
3 = A(x + 3) + B(x - 1)

Let x = 1: 3 = 4A → A = 3/4
Let x = -3: 3 = -4B → B = -3/4

∫ 3/[(x-1)(x+3)] dx = ∫ [3/4/(x-1) - 3/4/(x+3)] dx
= (3/4)ln|x-1| - (3/4)ln|x+3| + c
= (3/4)ln|x-1|/|x+3| + c

Example 2: Find ∫ (2x + 3)/[(x + 1)(x + 2)] dx

Solution:
(2x+3)/[(x+1)(x+2)] = A/(x+1) + B/(x+2)
2x + 3 = A(x + 2) + B(x + 1)

Let x = -1: 1 = A → A = 1
Let x = -2: -1 = -B → B = 1

∫ (2x+3)/[(x+1)(x+2)] dx = ∫ [1/(x+1) + 1/(x+2)] dx
= ln|x+1| + ln|x+2| + c
= ln|(x+1)(x+2)| + c

Important: If degree of numerator ≥ degree of denominator, divide first using polynomial long division!

8.6 Integration by Parts

The Reverse of Product Rule:
Used when integrating a product of two functions.

KEY FORMULA:

∫ u(dv/dx) dx = uv - ∫ v(du/dx) dx

Or simply: ∫ u dv = uv - ∫ v du

LIATE Rule for Choosing u

Choose u as the function that appears FIRST in this list:

• Logarithmic (ln x, log x)
• Inverse trig (tan⁻¹x, sin⁻¹x)
• Algebraic (x², x, √x)
• Trigonometric (sin x, cos x)
• Exponential (eˣ, e²ˣ)

Example 1: Find ∫ x eˣ dx

Solution:
u = x (algebraic), dv/dx = eˣ
du/dx = 1, v = eˣ

∫ x eˣ dx = x × eˣ - ∫ eˣ × 1 dx
= x eˣ - eˣ + c
= eˣ(x - 1) + c

Example 2: Find ∫ ln(x) dx

Solution:
Write as: ∫ 1 × ln(x) dx
u = ln(x), dv/dx = 1
du/dx = 1/x, v = x

∫ ln(x) dx = x ln(x) - ∫ x × (1/x) dx
= x ln(x) - ∫ 1 dx
= x ln(x) - x + c
= x(ln(x) - 1) + c

Example 3: Find ∫ x sin(x) dx

Solution:
u = x, dv/dx = sin(x)
du/dx = 1, v = -cos(x)

∫ x sin(x) dx = -x cos(x) - ∫ (-cos(x)) dx
= -x cos(x) + sin(x) + c

Example 4: Find ∫ x² eˣ dx (requires TWO applications!)

Solution:
First application:
u = x², dv/dx = eˣ
du/dx = 2x, v = eˣ
∫ x² eˣ dx = x² eˣ - 2∫ x eˣ dx

Second application on ∫ x eˣ dx:
= x eˣ - eˣ

Final:
∫ x² eˣ dx = x² eˣ - 2(x eˣ - eˣ) + c
= eˣ(x² - 2x + 2) + c

Example 5 (Definite): Evaluate ∫₀^π x sin(x) dx

Solution:
From Example 3: ∫ x sin(x) dx = -x cos(x) + sin(x) + c

= [-x cos(x) + sin(x)]₀^π
= [-π(-1) + 0] - [0 + 0]
= π

8.7 Combining Techniques

Strategy Guide

1. Always simplify first!

• Expand brackets
• Complete the square
• Factor if possible
• Divide if needed

2. Look for patterns:

• f'(x)/f(x) → ln|f(x)|
• 1/(x² + a²) → (1/a)tan⁻¹(x/a)
• Products → Integration by parts
• Composite functions → Substitution
• Rational functions → Partial fractions

Example 1: Find ∫ (3x + 7)/(x² + 2x + 5) dx

Solution:
Write: 3x + 7 = A(2x + 2) + B
3 = 2A → A = 3/2
7 = 2A + B → B = 4

= (3/2)∫ (2x+2)/(x²+2x+5) dx + 4∫ 1/(x²+2x+5) dx

First part: (3/2)ln(x² + 2x + 5)

Second part - complete square: x² + 2x + 5 = (x+1)² + 4
4∫ 1/[(x+1)² + 4] dx = 2tan⁻¹((x+1)/2)

Final: (3/2)ln(x² + 2x + 5) + 2tan⁻¹((x+1)/2) + c

Example 2: Find ∫ x³/(x² + 1) dx

Solution:
Polynomial division: x³/(x² + 1) = x - x/(x² + 1)

∫ x³/(x² + 1) dx = ∫ x dx - ∫ x/(x² + 1) dx
= x²/2 - (1/2)ln(x² + 1) + c

📋 Formula Quick Reference

Differentiation

d/dx[tan⁻¹(x)] = 1/(1 + x²)
d/dx[tan⁻¹(f(x))] = f'(x)/(1 + [f(x)]²)

Integration - Standard Forms

∫ 1/(1 + x²) dx = tan⁻¹(x) + c
∫ 1/(x² + a²) dx = (1/a)tan⁻¹(x/a) + c
∫ f'(x)/f(x) dx = ln|f(x)| + c

Integration - Methods

By Parts: ∫ u dv = uv - ∫ v du
Substitution: Replace dx with (dx/du)du
Partial Fractions: Split P(x)/Q(x) into simpler terms

Final Exam Tips:

• Don't forget '+ c' for indefinite integrals!
• Change limits when using substitution for definite integrals.
• Check if angles are in Radians (Calculus almost always uses Radians).
• If you get stuck, try a different method - if Substitution fails, try Parts!