FAC1004 Exam Focus — Leak Topics

Source: Leaked exam tips from lecturer session. Sir Hisham's question style — modified slightly ("diubah sikit"), not too hard but "mencabar sikit". Tangent expansion in Section B. Doing tutorials + lectures should be sufficient.

Quick Navigation

Topic Importance Section
Tangent Expansion (Taylor/Maclaurin) 🔴 CRITICAL — Section B #1. Tangent Expansion & Power Series (Section B)
Complex Numbers & De Moivre's Theorem 🟠 High #2. Complex Numbers & De Moivre's Theorem
Inverse Trigonometric Functions 🟠 High #3. Inverse Trigonometric Functions
Hyperbolic Functions 🟠 High #4. Hyperbolic Functions
Inverse Hyperbolic Functions 🟡 Medium #5. Inverse Hyperbolic Functions
Integration Involving Hyperbolic/Inverse Trig 🟡 Medium #6. Integration Techniques
Revision Checklist — #✅ Revision Checklist
Quick Reference Tables — #📋 Quick Reference Tables

1. Tangent Expansion & Power Series (Section B)

🔴 LEAK ALERT: This is explicitly named as a Section B topic. Practice expanding $\tan x$ as a Maclaurin series up to $x^5$ or $x^7$ — both by direct differentiation AND by division of known series.

1.1 Taylor & Maclaurin Series — Refresher

Taylor series of $f(x)$ about $x = a$:

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$$

Maclaurin series (special case, $a = 0$):

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

Key point: "Tangent expansion" almost certainly means finding the Maclaurin series for $\tan x$, or possibly $\tan^{-1} x$ (arctan) — but given the leak says "Tangen expansion(?)" with a question mark, the most likely candidate is $\tan x$.

1.2 Standard Maclaurin Series You MUST Know

Function Series Valid For
$e^x$ $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ all $x$
$\sin x$ $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$ all $x$
$\cos x$ $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$ all $x$
$\ln(1+x)$ $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots$ $-1 < x \leq 1$
$\arctan x$ $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$ $-1 \leq x \leq 1$
$\frac{1}{1-x}$ $\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$ $
$\frac{1}{1+x}$ $\displaystyle\sum_{n=0}^{\infty} (-1)^n x^n = 1 - x + x^2 - x^3 + \cdots$ $

1.3 Expanding $\tan x$ — Method 1: Direct Differentiation

Compute derivatives of $f(x) = \tan x$ at $x = 0$:

$n$ $f^{(n)}(x)$ $f^{(n)}(0)$
0 $\tan x$ $0$
1 $\sec^2 x$ $1$
2 $2\sec^2 x \tan x$ $0$
3 $2\sec^4 x + 4\sec^2 x \tan^2 x$ $2$
4 $16\sec^4 x \tan x + 8\sec^2 x \tan^3 x$ $0$
5 $16\sec^6 x + 64\sec^4 x \tan^2 x + 16\sec^2 x \tan^4 x$ $16$

Derivative details (step-by-step):

$$f'(x) = \frac{d}{dx}\tan x = \sec^2 x \quad\Rightarrow\quad f'(0) = \sec^2 0 = 1$$

$$f''(x) = \frac{d}{dx}\sec^2 x = 2\sec x \cdot \sec x \tan x = 2\sec^2 x \tan x \quad\Rightarrow\quad f''(0) = 0$$

$$f'''(x) = \frac{d}{dx}[2\sec^2 x \tan x] = 2[2\sec^2 x \tan x \cdot \tan x + \sec^2 x \cdot \sec^2 x]$$ $$= 2[2\sec^2 x \tan^2 x + \sec^4 x] = 4\sec^2 x \tan^2 x + 2\sec^4 x$$ $$\Rightarrow\quad f'''(0) = 4(1)(0) + 2(1) = 2$$

$$f^{(4)}(x) = \frac{d}{dx}[4\sec^2 x \tan^2 x + 2\sec^4 x] \quad\Rightarrow\quad f^{(4)}(0) = 0$$

$$f^{(5)}(0) = 16$$

Therefore:

$$\boxed{\tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + \cdots}$$

Tip: For exam purposes, you only need up to $x^3$ or $x^5$. Memorise at minimum: $\tan x \approx x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$

1.4 Expanding $\tan x$ — Method 2: Division of Known Series ($\tan x = \frac{\sin x}{\cos x}$)

This is often easier and less error-prone than direct differentiation!

$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \cdots$$

$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \cdots$$

Perform polynomial long division:

$$ \require{enclose} \begin{array}{r} x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \cdots \[-3pt] 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots \enclose{longdiv}{;x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \cdots} \[-3pt] \underline{x - \frac{x^3}{2} + \frac{x^5}{24} - \frac{x^7}{720} + \cdots} \[-3pt] \left(\frac{1}{3}\right)x^3 + \left(\frac{1}{120} - \frac{1}{24}\right)x^5 + \cdots \[-3pt] \underline{\frac{1}{3}x^3 - \frac{1}{6}x^5 + \cdots} \[-3pt] \frac{2}{15}x^5 + \cdots \end{array} $$

This confirms: $$\tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \cdots$$

1.5 Other Related Expansions You Should Know

Series for $\sec x$ (derived from $1/\cos x$):

$$\sec x = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \frac{61x^6}{720} + \cdots$$

Series for $\arctan x$:

$$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$

Since $\tan x$ is in Section B, Sir Hisham's style may involve:

  • Finding the Maclaurin series up to a given term
  • Using the series to estimate $\tan x$ at a small value
  • Differentiating the series term-by-term
  • Relating $\tan x$ to series for $\sin x$, $\cos x$

1.6 Substitution & Multiplication Techniques

Substitution: Replace $x$ with $g(x)$ in a known series.

Example: Find series for $e^{-x^2}$ up to $x^4$: $$e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$$ Let $u = -x^2$: $$e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots$$

Multiplication: Multiply two known series term-by-term.

Example: Find series for $x\sin x$ up to $x^5$: $$x\sin x = x\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right) = x^2 - \frac{x^4}{6} + \frac{x^6}{120} - \cdots$$

2. Complex Numbers & De Moivre's Theorem

2.1 Essential Formulas

Concept Formula
Cartesian form $z = a + bi$
Polar form $z = r(\cos\theta + i\sin\theta)$
Exponential form $z = re^{i\theta}$
Modulus $
Argument $\arg(z) = \tan^{-1}(b/a)$ (watch quadrant!)
Complex conjugate $\bar{z} = a - bi$, $z\bar{z} =
Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$

2.2 De Moivre's Theorem

$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$

For powers: $$z^n = r^n(\cos n\theta + i\sin n\theta) = r^n e^{in\theta}$$

For $n$-th roots: $$z^{1/n} = r^{1/n}\left[\cos\left(\frac{\theta + 2\pi k}{n}\right) + i\sin\left(\frac{\theta + 2\pi k}{n}\right)\right], \quad k = 0, 1, \ldots, n-1$$

Roots of unity: $$z^n = 1 \quad\Rightarrow\quad z = e^{2\pi i k/n}, \quad k = 0, 1, \ldots, n-1$$

2.3 Argand Diagram — Geometric Loci

Locus Equation Description
Circle $ z - z_0
Disc $ z - z_0
Perp bisector $ z - z_1
Half-line $\arg(z - z_0) = \theta$ Ray from $z_0$ at angle $\theta$
Argand interval $\alpha \leq \arg(z - z_0) \leq \beta$ Angular sector

2.4 Complex Logarithms

$$\text{Log}(z) = \ln|z| + i\arg(z) \quad \text{(principal branch, } -\pi < \arg(z) \leq \pi\text{)}$$

$$\log(z) = \ln|z| + i(\arg(z) + 2\pi k), \quad k \in \mathbb{Z}$$

$$\ln(z_1 z_2) = \ln z_1 + \ln z_2 + 2\pi i k \quad \text{(branch cut issues!)}$$

2.5 Sir Hisham-Style Problem Patterns

mindmap
  root((Complex Numbers<br/>Exam Patterns))
    De Moivre Powers
      Express cos(nθ) in powers of cos θ
      Express sin(nθ) in powers of sin θ
      Find exact trig values
    Roots of Complex Numbers
      Find cube roots of a+bi
      Plot on Argand diagram
      Roots of unity applications
    Geometric Loci
      Sketch |z-1| = |z+i|
      Find region: |z| ≤ 2, 0 ≤ arg(z) ≤ π/4
      Intersection of two loci
    Complex Logarithm
      Compute Log(z) for given z
      Solve equations with Log
      Branch cut awareness
    Euler's Formula
      Prove trig identities
      Express cos⁴θ in terms of multiple angles
      Sum trig series

3. Inverse Trigonometric Functions

3.1 Definitions & Domains

Function Domain Range (Principal) Derivative
$\sin^{-1} x$ $[-1, 1]$ $[-\pi/2, \pi/2]$ $\frac{1}{\sqrt{1-x^2}}$
$\cos^{-1} x$ $[-1, 1]$ $[0, \pi]$ $-\frac{1}{\sqrt{1-x^2}}$
$\tan^{-1} x$ $\mathbb{R}$ $(-\pi/2, \pi/2)$ $\frac{1}{1+x^2}$
$\cot^{-1} x$ $\mathbb{R}$ $(0, \pi)$ $-\frac{1}{1+x^2}$
$\sec^{-1} x$ $(-\infty,-1]\cup[1,\infty)$ $[0,\pi/2)\cup(\pi/2,\pi]$ $\frac{1}{
$\csc^{-1} x$ $(-\infty,-1]\cup[1,\infty)$ $[-\pi/2,0)\cup(0,\pi/2]$ $-\frac{1}{

3.2 Key Identities

Complementary:

  • $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$
  • $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$
  • $\sec^{-1} x + \csc^{-1} x = \frac{\pi}{2}$

Sum/Difference:

  • $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$
  • $\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$

Double Angle:

  • $2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$

Interconversion:

  • $\sin^{-1} x = \cos^{-1}(\sqrt{1-x^2}) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$

3.3 Derivative Patterns (Chain Rule)

$f(x)$ $f'(x)$
$\sin^{-1}(g(x))$ $\frac{g'(x)}{\sqrt{1-[g(x)]^2}}$
$\cos^{-1}(g(x))$ $\frac{-g'(x)}{\sqrt{1-[g(x)]^2}}$
$\tan^{-1}(g(x))$ $\frac{g'(x)}{1+[g(x)]^2}$

4. Hyperbolic Functions

4.1 Definitions (Exponential Form)

$$\sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}, \qquad \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$

Relationship to trig (Osborne's rule): Replace $\cos \to \cosh$, $\sin \to \sinh$, and flip sign of any product of two sines.

4.2 Fundamental Identity

$$\cosh^2 x - \sinh^2 x = 1$$

4.3 Other Key Identities

  • $\sinh(2x) = 2\sinh x \cosh x$
  • $\cosh(2x) = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1 = 2\sinh^2 x + 1$
  • $\sinh(x \pm y) = \sinh x \cosh y \pm \cosh x \sinh y$
  • $\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y$
  • $\cosh x + \sinh x = e^x$
  • $\cosh x - \sinh x = e^{-x}$
  • $1 - \tanh^2 x = \text{sech}^2 x$
  • $\coth^2 x - 1 = \text{csch}^2 x$

4.4 Derivatives

$f(x)$ $f'(x)$
$\sinh x$ $\cosh x$
$\cosh x$ $\sinh x$
$\tanh x$ $\text{sech}^2 x$
$\coth x$ $-\text{csch}^2 x$

4.5 Basic Integrals

  • $\int \sinh x , dx = \cosh x + C$
  • $\int \cosh x , dx = \sinh x + C$
  • $\int \text{sech}^2 x , dx = \tanh x + C$
  • $\int \tanh x , dx = \ln(\cosh x) + C$

5. Inverse Hyperbolic Functions

5.1 Logarithmic Forms

$$\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1}), \quad \text{domain: } \mathbb{R}$$

$$\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1}), \quad \text{domain: } x \geq 1$$

$$\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right), \quad \text{domain: } |x| < 1$$

$$\coth^{-1} x = \frac{1}{2}\ln\left(\frac{x+1}{x-1}\right), \quad \text{domain: } |x| > 1$$

5.2 Derivatives

$f(x)$ $f'(x)$ Domain
$\sinh^{-1} x$ $\frac{1}{\sqrt{1+x^2}}$ all $x$
$\cosh^{-1} x$ $\frac{1}{\sqrt{x^2-1}}$ $x > 1$
$\tanh^{-1} x$ $\frac{1}{1-x^2}$ $
$\coth^{-1} x$ $\frac{1}{1-x^2}$ $

⚠️ Note: $\tanh^{-1}$ and $\coth^{-1}$ have the same derivative formula but different domains!

5.3 Integrals Leading to Inverse Hyperbolic Functions

Integral Result Condition
$\int \frac{dx}{\sqrt{a^2 + x^2}}$ $\sinh^{-1}\left(\frac{x}{a}\right) + C$ all $x$
$\int \frac{dx}{\sqrt{x^2 - a^2}}$ $\cosh^{-1}\left(\frac{x}{a}\right) + C$ $x > a$
$\int \frac{dx}{a^2 - x^2}$ $\frac{1}{a}\tanh^{-1}\left(\frac{x}{a}\right) + C$ $
$\int \frac{dx}{a^2 - x^2}$ $\frac{1}{a}\coth^{-1}\left(\frac{x}{a}\right) + C$ $

6. Integration Techniques

6.1 Hyperbolic Substitution

Radical Form Substitution Differential
$\sqrt{a^2 + x^2}$ $x = a\sinh u$ $dx = a\cosh u , du$
$\sqrt{x^2 - a^2}$ $x = a\cosh u$ $dx = a\sinh u , du$
$\sqrt{a^2 - x^2}$ $x = a\sin\theta$ (trig) —

6.2 Integration by Parts with Inverse Trig/Hyperbolic

Typical pattern: integrate functions like $\sin^{-1} x$, $\tan^{-1} x$, $\sinh^{-1} x$ using:

$$\int \sin^{-1} x , dx = x\sin^{-1} x + \sqrt{1-x^2} + C$$

$$\int \tan^{-1} x , dx = x\tan^{-1} x - \frac{1}{2}\ln(1+x^2) + C$$

$$\int \sinh^{-1} x , dx = x\sinh^{-1} x - \sqrt{1+x^2} + C$$

6.3 $u$-Substitution with Hyperbolic Functions

Substitution When to use
$u = \sinh x$ $\cosh x , dx$ present
$u = \cosh x$ $\sinh x , dx$ present
$u = \tanh x$ $\text{sech}^2 x , dx$ present
$u = \coth x$ $\text{csch}^2 x , dx$ present

7. Section B — Extended Practice Problems

These are modelled after Sir Hisham's question style (modified slightly, "mencabar sikit") with emphasis on tangent expansion.

Problem B1: Maclaurin Series of $\tan x$ (Direct & Division Methods)

(a) By differentiating $f(x) = \tan x$, find $f'(x)$, $f''(x)$, $f'''(x)$, $f^{(4)}(x)$, and $f^{(5)}(x)$. Hence find the Maclaurin series expansion of $\tan x$ up to the term in $x^5$.

(b) Verify your answer by using the series for $\sin x$ and $\cos x$ and performing the division $\tan x = \frac{\sin x}{\cos x}$.

(c) Use your series to estimate $\tan(0.2)$ and compare with the exact value.

Solution Outline

(a) Using derivatives:

  • $f(0) = 0$, $f'(0) = 1$, $f''(0) = 0$, $f'''(0) = 2$, $f^{(4)}(0) = 0$, $f^{(5)}(0) = 16$

$$\tan x = x + \frac{2}{3!}x^3 + \frac{16}{5!}x^5 + \cdots = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \cdots$$

(b) Division method: $\frac{x - x^3/6 + x^5/120 - \cdots}{1 - x^2/2 + x^4/24 - \cdots} = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots$ ✓

(c) $\tan(0.2) \approx 0.2 + \frac{0.008}{3} + \frac{2(0.00032)}{15} = 0.2 + 0.002667 + 0.0000427 \approx 0.20271$ Exact: $\tan(0.2) \approx 0.20271$ — series matches well!

Problem B2: Series for $\sec x$ and Relation to $\tan x$

(a) Using $1/\cos x$, find the Maclaurin series for $\sec x$ up to $x^4$.

(b) Verify that $\frac{d}{dx}\tan x = \sec^2 x$ by differentiating your series from Problem B1 and squaring your series for $\sec x$.

Solution Outline

(a) $\sec x = 1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$

(b) $\frac{d}{dx}\left(x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots\right) = 1 + x^2 + \frac{2x^4}{3} + \cdots$ And $\sec^2 x = \left(1 + \frac{x^2}{2} + \frac{5x^4}{24}\right)^2 = 1 + x^2 + \frac{2x^4}{3} + \cdots$ ✓

Problem B3: Complex Numbers + De Moivre — Modified from Tutorial

(a) Express $z = -1 + \sqrt{3},i$ in polar form $re^{i\theta}$.

(b) Use De Moivre's theorem to find $z^5$, expressing your answer in Cartesian form $a + bi$.

(c) Find all cube roots of $z$ and plot them on an Argand diagram.

(d) Solve the equation $w^3 = -1 + \sqrt{3},i$ and express each root in the form $re^{i\theta}$.

Solution Outline

(a) $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$, $\theta = \tan^{-1}(\sqrt{3}/-1) = \tan^{-1}(-\sqrt{3}) = \frac{2\pi}{3}$ (Quadrant II) So $z = 2e^{i(2\pi/3)}$.

(b) $z^5 = 2^5 e^{i(10\pi/3)} = 32\left(\cos\frac{10\pi}{3} + i\sin\frac{10\pi}{3}\right)$ $\frac{10\pi}{3} = \frac{4\pi}{3} + 2\pi$, so $\cos\frac{4\pi}{3} = -\frac{1}{2}$, $\sin\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$ $$z^5 = 32\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -16 - 16\sqrt{3},i$$

(c) Cube roots: $z^{1/3} = 2^{1/3}e^{i(2\pi/3 + 2\pi k)/3}$ for $k = 0,1,2$ $k=0$: $2^{1/3}e^{i(2\pi/9)}$, $k=1$: $2^{1/3}e^{i(8\pi/9)}$, $k=2$: $2^{1/3}e^{i(14\pi/9)}$

(d) Same as (c) — $w = z^{1/3}$.

Problem B4: Inverse Trig & Hyperbolic — Mixed

(a) Differentiate $y = \sin^{-1}(e^{2x})$ and simplify.

(b) Show that $\tanh^{-1} x + \tanh^{-1} y = \tanh^{-1}\left(\frac{x+y}{1+xy}\right)$.

(c) Evaluate $\int \frac{dx}{\sqrt{9 + x^2}}$ using hyperbolic substitution $x = 3\sinh u$.

Solution Outline

(a) $\frac{dy}{dx} = \frac{2e^{2x}}{\sqrt{1 - e^{4x}}}$

(b) Let $\tanh^{-1} x = a$, $\tanh^{-1} y = b$. Then $x = \tanh a$, $y = \tanh b$. $$\tanh(a+b) = \frac{\tanh a + \tanh b}{1 + \tanh a \tanh b} = \frac{x+y}{1+xy}$$ $$\therefore a+b = \tanh^{-1}\left(\frac{x+y}{1+xy}\right)$$

(c) Let $x = 3\sinh u$, $dx = 3\cosh u , du$. $$\int \frac{3\cosh u}{\sqrt{9 + 9\sinh^2 u}} du = \int \frac{3\cosh u}{3\sqrt{1+\sinh^2 u}} du = \int \frac{\cosh u}{\cosh u} du = \int du = u + C$$ $$= \sinh^{-1}\left(\frac{x}{3}\right) + C = \ln\left(\frac{x + \sqrt{x^2+9}}{3}\right) + C$$

Problem B5: Integration of Hyperbolic Functions — Tutorial Style

Evaluate the following integrals:

(a) $\displaystyle\int \sinh(3x) , dx$

(b) $\displaystyle\int_0^1 \cosh(2x) , dx$

(c) $\displaystyle\int \tanh x , dx$

(d) $\displaystyle\int \frac{dx}{\sqrt{x^2 - 4}}$ for $x > 2$

Solution Outline

(a) $\int \sinh(3x) dx = \frac{1}{3}\cosh(3x) + C$

(b) $\int_0^1 \cosh(2x) dx = \left[\frac{1}{2}\sinh(2x)\right]_0^1 = \frac{1}{2}\sinh(2)$

(c) $\int \tanh x , dx = \int \frac{\sinh x}{\cosh x} dx = \ln(\cosh x) + C$

(d) $\int \frac{dx}{\sqrt{x^2-4}} = \cosh^{-1}\left(\frac{x}{2}\right) + C = \ln\left|x + \sqrt{x^2-4}\right| + C$

✅ Revision Checklist

Before the exam, tick off each item:

Power Series & Tangent Expansion 🎯

  • [ ] I know the Taylor series formula $f(x) = \sum \frac{f^{(n)}(a)}{n!}(x-a)^n$
  • [ ] I know the Maclaurin series formula $f(x) = \sum \frac{f^{(n)}(0)}{n!}x^n$
  • [ ] I can recite the standard series for $e^x$, $\sin x$, $\cos x$, $\ln(1+x)$, $\arctan x$
  • [ ] I can find the Maclaurin series for $\tan x$ by direct differentiation up to $x^5$
  • [ ] I can find the Maclaurin series for $\tan x$ by division of $\sin x / \cos x$
  • [ ] I can apply substitution (e.g., $e^{-x^2}$, $\sin(x^2)$)
  • [ ] I can apply multiplication of series (e.g., $x\sin x$, $e^x\cos x$)
  • [ ] I understand radius of convergence (basic)

Complex Numbers 🎯

  • [ ] I can convert between Cartesian, polar, and exponential forms
  • [ ] I can apply De Moivre's theorem for powers and roots
  • [ ] I can find $n$-th roots of any complex number
  • [ ] I can sketch loci (circles, perpendicular bisectors, half-lines)
  • [ ] I know Euler's formula and its applications
  • [ ] I can compute complex logarithms (principal value)
  • [ ] I know how to express $\cos^n\theta$, $\sin^n\theta$ in multiple angles

Inverse Trig Functions 🎯

  • [ ] I know all 6 definitions, domains, and ranges
  • [ ] I can differentiate any inverse trig function (with chain rule)
  • [ ] I know $\tan^{-1}$ sum/difference and double-angle formulas
  • [ ] I can interconvert between inverse trig functions
  • [ ] I can simplify expressions like $\sin(\cos^{-1} x)$

Hyperbolic Functions 🎯

  • [ ] I know the exponential definitions of $\sinh x$, $\cosh x$, $\tanh x$
  • [ ] I know $\cosh^2 x - \sinh^2 x = 1$ and derived identities
  • [ ] I know double-angle formulas
  • [ ] I can differentiate and integrate hyperbolic functions
  • [ ] I can perform hyperbolic substitution ($x = a\sinh u$, $x = a\cosh u$)

Inverse Hyperbolic Functions 🎯

  • [ ] I know logarithmic forms of $\sinh^{-1}$, $\cosh^{-1}$, $\tanh^{-1}$
  • [ ] I know their derivatives and domain restrictions
  • [ ] I can integrate using standard inverse hyperbolic forms
  • [ ] I know $\frac{d}{dx}\tanh^{-1}x = \frac{1}{1-x^2}$ and $\frac{d}{dx}\coth^{-1}x = \frac{1}{1-x^2}$ — same formula, different domains!

📋 Quick Reference Tables

Derivatives Cheat Sheet

Function Derivative
$\tan x$ $\sec^2 x$
$\sec x$ $\sec x \tan x$
$\sin^{-1} x$ $1/\sqrt{1-x^2}$
$\cos^{-1} x$ $-1/\sqrt{1-x^2}$
$\tan^{-1} x$ $1/(1+x^2)$
$\sinh x$ $\cosh x$
$\cosh x$ $\sinh x$
$\tanh x$ $\text{sech}^2 x$
$\sinh^{-1} x$ $1/\sqrt{1+x^2}$
$\cosh^{-1} x$ $1/\sqrt{x^2-1}$
$\tanh^{-1} x$ $1/(1-x^2)$, $

Integrals Cheat Sheet

Integral Result
$\int \sec^2 x , dx$ $\tan x + C$
$\int \tan x , dx$ $\ln|\sec x| + C$
$\int \sinh x , dx$ $\cosh x + C$
$\int \cosh x , dx$ $\sinh x + C$
$\int \tanh x , dx$ $\ln(\cosh x) + C$
$\int \frac{dx}{\sqrt{a^2+x^2}}$ $\sinh^{-1}(x/a) + C$
$\int \frac{dx}{\sqrt{x^2-a^2}}$ $\cosh^{-1}(x/a) + C$
$\int \frac{dx}{a^2-x^2}$ $\frac{1}{a}\tanh^{-1}(x/a) + C$ ($|x|<a$)
$\int \frac{dx}{1+x^2}$ $\tan^{-1} x + C$
$\int \frac{dx}{\sqrt{1-x^2}}$ $\sin^{-1} x + C$

Common Maclaurin Series

Function Up to $x^5$
$\tan x$ $x + \dfrac{x^3}{3} + \dfrac{2x^5}{15} + \cdots$
$\sin x$ $x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$
$\cos x$ $1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$
$\sec x$ $1 + \dfrac{x^2}{2} + \dfrac{5x^4}{24} + \cdots$
$\arctan x$ $x - \dfrac{x^3}{3} + \dfrac{x^5}{5} - \cdots$
$e^x$ $1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \dfrac{x^5}{120} + \cdots$

📝 Exam Day Reminders

  1. Section B is where the tangent expansion lives — allocate extra time here
  2. Sketch Argand diagrams for loci questions — partial marks even if algebra is wrong
  3. For inverse trig/hyperbolic: state domain restrictions to show you understand branch cuts
  4. For Maclaurin series of $\tan x$: both methods work — use whichever is faster (division is usually quicker!)
  5. If stuck on a series problem: start from $e^x$, $\sin x$, $\cos x$ and manipulate
  6. Integration with hyperbolic substitution: don't forget to back-substitute $u$ to $x$
  7. Tutorial questions are the blueprint — Sir Hisham modifies from these, so if you've done all tutorials you've seen the core ideas

🔗 Related Pages

Built from leaked exam tips. Double-check against your own lecture notes and tutorials. Good luck! 🍀