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A-Level Further Mathematics: Further Pure 2

A-Level Further Mathematics · Edexcel / AQA / OCR / CIE · Updated March 2026 · 40 min read

Exam Board Note

Further Pure 2 content is found in Edexcel FP2 (8FM0:02), AQA Further Mathematics options, OCR A Further Pure B (Y542), and CIE Further Mathematics Paper 2. Polar coordinates and hyperbolic functions appear on all boards; the t-substitution is particularly prominent on Edexcel. Check your specification for exact topic coverage, as some boards integrate these topics across different papers.

Further Pure 2 extends the calculus and analysis of Further Pure 1. You will encounter elegant new coordinate systems, an analogy of the trigonometric functions built from exponentials, and advanced techniques that appear in undergraduate mathematics. Many of the techniques here appear in A-Level examinations as multi-part questions worth 8–12 marks.

1. Polar Coordinates and Curves

Definition: Polar Coordinates

A point $P$ in the plane is described by polar coordinates $(r, \theta)$ where $r \geq 0$ is the distance from the origin (the pole) and $\theta$ is the angle from the positive $x$-axis (the initial line), measured anticlockwise.

Conversion between Cartesian and polar:

$$x = r\cos\theta, \quad y = r\sin\theta \qquad \longleftrightarrow \qquad r = \sqrt{x^2+y^2}, \quad \tan\theta = \frac{y}{x}$$

Common Polar Curves

You should be able to recognise and sketch the following families of polar curves:

Worked Example 1.1 — Sketching a Polar Curve

Sketch the cardioid $r = 2(1 + \cos\theta)$ for $0 \leq \theta \leq 2\pi$ and state its Cartesian equation.

Step 1 Tabulate key values:

$\theta = 0$: $r = 4$  |  $\theta = \frac{\pi}{2}$: $r = 2$  |  $\theta = \pi$: $r = 0$  |  $\theta = \frac{3\pi}{2}$: $r = 2$  |  $\theta = 2\pi$: $r = 4$

The curve passes through the origin at $\theta = \pi$, has maximum radius 4 along the positive $x$-axis, and is symmetric about the initial line.

Step 2 For the Cartesian equation, note $r = 2 + 2\cos\theta = 2 + 2\cdot\dfrac{x}{r}$, so $r^2 = 2r + 2x$, giving: $$x^2 + y^2 = 2\sqrt{x^2+y^2} + 2x$$ $$(x^2+y^2-2x)^2 = 4(x^2+y^2)$$

FP2·1 — Polar curves: cardioid $r = 2(1+\cos\theta)$ (blue), four-petal rose $r = 3\cos(2\theta)$ (purple), and circle $r = 2$ (green). The cardioid passes through the origin at $\theta = \pi$ and reaches maximum $r = 4$ at $\theta = 0$.

2. Area in Polar Coordinates

Area Formula in Polar Coordinates

The area enclosed by the polar curve $r = f(\theta)$ between $\theta = \alpha$ and $\theta = \beta$ is:

$$A = \frac{1}{2}\int_{\alpha}^{\beta} r^2 \, d\theta$$

This formula counts the area swept out by the radius vector from angle $\alpha$ to $\beta$.

Worked Example 2.1 — Area of a Cardioid

Find the area enclosed by the cardioid $r = a(1 + \cos\theta)$.

Step 1 The cardioid is traced once as $\theta$ goes from $0$ to $2\pi$. By symmetry about $\theta = 0$: $$A = \frac{1}{2}\int_0^{2\pi} a^2(1+\cos\theta)^2\,d\theta = a^2\int_0^{\pi}(1+\cos\theta)^2\,d\theta$$

Step 2 Expand $(1+\cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta = 1 + 2\cos\theta + \dfrac{1+\cos 2\theta}{2}$.

Step 3 Integrate: $$\int_0^{\pi}\!\left(\frac{3}{2} + 2\cos\theta + \frac{\cos 2\theta}{2}\right)d\theta = \left[\frac{3\theta}{2} + 2\sin\theta + \frac{\sin 2\theta}{4}\right]_0^{\pi}$$ $$= \frac{3\pi}{2} + 0 + 0 - 0 = \frac{3\pi}{2}$$

Result $A = a^2 \cdot \dfrac{3\pi}{2} = \dfrac{3\pi a^2}{2}$.

Worked Example 2.2 — Area Between Two Polar Curves

Find the area of the region inside $r = 3\cos\theta$ but outside $r = 1 + \cos\theta$.

Step 1 Find intersection: $3\cos\theta = 1+\cos\theta$, so $2\cos\theta = 1$, giving $\theta = \pm\dfrac{\pi}{3}$.

Step 2 The required area (using symmetry): $$A = 2 \cdot \frac{1}{2}\int_0^{\pi/3}\!\left[(3\cos\theta)^2 - (1+\cos\theta)^2\right]d\theta$$ $$= \int_0^{\pi/3}(9\cos^2\theta - 1 - 2\cos\theta - \cos^2\theta)\,d\theta$$ $$= \int_0^{\pi/3}(8\cos^2\theta - 2\cos\theta - 1)\,d\theta$$

Step 3 Use $\cos^2\theta = \dfrac{1+\cos 2\theta}{2}$: $$= \int_0^{\pi/3}\!\left(4 + 4\cos 2\theta - 2\cos\theta - 1\right)d\theta = \left[3\theta + 2\sin 2\theta - 2\sin\theta\right]_0^{\pi/3}$$ $$= \pi + 2\cdot\frac{\sqrt{3}}{2}\cdot 2 - 2\cdot\frac{\sqrt{3}}{2} = \pi + 2\sqrt{3} - \sqrt{3} = \pi + \sqrt{3}$$

3. Hyperbolic Functions

The hyperbolic functions are defined using exponentials. They satisfy identities that closely mirror the trigonometric identities, but arise from the hyperbola $x^2 - y^2 = 1$ rather than the unit circle.

Definitions

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

Also: $\text{sech}\,x = \dfrac{1}{\cosh x}$, $\text{cosech}\,x = \dfrac{1}{\sinh x}$, $\text{coth}\,x = \dfrac{1}{\tanh x}$.

Key differences from trigonometric functions:

Derivatives: $$\frac{d}{dx}\sinh x = \cosh x, \quad \frac{d}{dx}\cosh x = \sinh x, \quad \frac{d}{dx}\tanh x = \text{sech}^2 x$$

Note: the derivative of $\cosh x$ is $\sinh x$ (not $-\sinh x$, unlike $\cos x \to -\sin x$).

4. Hyperbolic Identities and Inverse Functions

Osborn's Rule

Osborn's Rule provides a systematic way to convert trigonometric identities to hyperbolic ones: replace every $\sin$ with $\sinh$, every $\cos$ with $\cosh$, and change the sign of every term containing a product of two $\sinh$ functions.

Key Hyperbolic Identities

$$\cosh^2 x - \sinh^2 x = 1 \quad \text{(fundamental identity)}$$

$$\sinh(x \pm y) = \sinh x \cosh y \pm \cosh x \sinh y$$

$$\cosh(x \pm y) = \cosh x \cosh y \pm \sinh x \sinh y$$

$$\sinh 2x = 2\sinh x \cosh x, \qquad \cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1$$

Inverse Hyperbolic Functions

The inverse hyperbolic functions can be expressed in terms of natural logarithms:

$$\sinh^{-1} x = \ln\!\left(x + \sqrt{x^2+1}\right), \quad x \in \mathbb{R}$$ $$\cosh^{-1} x = \ln\!\left(x + \sqrt{x^2-1}\right), \quad x \geq 1$$ $$\tanh^{-1} x = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right), \quad |x| < 1$$

Derivatives of inverse hyperbolic functions: $$\frac{d}{dx}\sinh^{-1}\!\left(\frac{x}{a}\right) = \frac{1}{\sqrt{x^2+a^2}}, \qquad \frac{d}{dx}\cosh^{-1}\!\left(\frac{x}{a}\right) = \frac{1}{\sqrt{x^2-a^2}}, \quad x > a$$

These lead to important integration results: $$\int \frac{1}{\sqrt{x^2+a^2}}\,dx = \sinh^{-1}\!\left(\frac{x}{a}\right) + C = \ln\!\left(x+\sqrt{x^2+a^2}\right)+C$$ $$\int \frac{1}{\sqrt{x^2-a^2}}\,dx = \cosh^{-1}\!\left(\frac{x}{a}\right) + C = \ln\!\left(x+\sqrt{x^2-a^2}\right)+C, \quad x > a$$

Worked Example 4.1 — Solving a Hyperbolic Equation

Solve $5\cosh x - 3\sinh x = 7$.

Step 1 Substitute the definitions $\cosh x = \dfrac{e^x + e^{-x}}{2}$ and $\sinh x = \dfrac{e^x - e^{-x}}{2}$: $$5\cdot\frac{e^x+e^{-x}}{2} - 3\cdot\frac{e^x - e^{-x}}{2} = 7$$ $$\frac{5e^x + 5e^{-x} - 3e^x + 3e^{-x}}{2} = 7$$ $$2e^x + 8e^{-x} = 14$$ $$e^x + 4e^{-x} = 7$$

Step 2 Multiply through by $e^x$: $e^{2x} - 7e^x + 4 = 0$ (quadratic in $e^x$): $$e^x = \frac{7 \pm \sqrt{49-16}}{2} = \frac{7 \pm \sqrt{33}}{2}$$ Both values are positive, so both give real solutions: $$x = \ln\!\left(\frac{7+\sqrt{33}}{2}\right) \approx 1.622 \quad \text{or} \quad x = \ln\!\left(\frac{7-\sqrt{33}}{2}\right) \approx -0.622$$

5. Reduction Formulae

A reduction formula is a recurrence relation connecting $I_n$ (an integral depending on a parameter $n$) to $I_{n-1}$ or $I_{n-2}$. They are derived by applying integration by parts strategically.

Worked Example 5.1 — Reduction Formula for $I_n = \int_0^{\pi/2}\sin^n x\,dx$

Show that $I_n = \dfrac{n-1}{n}I_{n-2}$ for $n \geq 2$, and use it to evaluate $I_5$.

Step 1 Write $\sin^n x = \sin^{n-1}x \cdot \sin x$ and integrate by parts with $u = \sin^{n-1}x$, $dv = \sin x\,dx$:

$u = \sin^{n-1}x \Rightarrow du = (n-1)\sin^{n-2}x\cos x\,dx$

$dv = \sin x\,dx \Rightarrow v = -\cos x$

$$I_n = \left[-\cos x\sin^{n-1}x\right]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\cos^2 x\sin^{n-2}x\,dx$$

Step 2 The boundary term is zero at both limits. Use $\cos^2 x = 1 - \sin^2 x$: $$I_n = (n-1)\int_0^{\pi/2}(1-\sin^2 x)\sin^{n-2}x\,dx = (n-1)(I_{n-2} - I_n)$$

Step 3 Solve for $I_n$: $I_n + (n-1)I_n = (n-1)I_{n-2}$, so: $$I_n = \frac{n-1}{n}I_{n-2}$$

Step 4 Apply repeatedly with $I_1 = \displaystyle\int_0^{\pi/2}\sin x\,dx = 1$: $$I_5 = \frac{4}{5}I_3 = \frac{4}{5}\cdot\frac{2}{3}I_1 = \frac{4}{5}\cdot\frac{2}{3}\cdot 1 = \frac{8}{15}$$

Worked Example 5.2 — Reduction Formula for $I_n = \int x^n e^x\,dx$

Find a reduction formula for $I_n = \displaystyle\int x^n e^x\,dx$.

Integrate by parts: $u = x^n$, $dv = e^x\,dx$, so $du = nx^{n-1}\,dx$, $v = e^x$: $$I_n = x^n e^x - n\int x^{n-1}e^x\,dx = x^n e^x - nI_{n-1}$$ With base case $I_0 = e^x + C$.

6. Arc Length and Surface of Revolution

Arc Length Formulae

For a Cartesian curve $y = f(x)$ from $x = a$ to $x = b$: $$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$$

For a parametric curve $x = x(t)$, $y = y(t)$ from $t = t_1$ to $t = t_2$: $$L = \int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$$

For a polar curve $r = f(\theta)$ from $\theta = \alpha$ to $\theta = \beta$: $$L = \int_{\alpha}^{\beta}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta$$

Surface of Revolution

When the curve $y = f(x)$, $a \leq x \leq b$, is rotated through $2\pi$ about the $x$-axis, the surface area is: $$S = 2\pi\int_a^b y\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$$

For rotation about the $y$-axis: $S = 2\pi\displaystyle\int_a^b x\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$

Worked Example 6.1 — Arc Length

Find the length of the curve $y = \cosh x$ from $x = 0$ to $x = \ln 2$.

Step 1 $\dfrac{dy}{dx} = \sinh x$, so $1 + \left(\dfrac{dy}{dx}\right)^2 = 1 + \sinh^2 x = \cosh^2 x$.

Step 2 $$L = \int_0^{\ln 2}\sqrt{\cosh^2 x}\,dx = \int_0^{\ln 2}\cosh x\,dx = \left[\sinh x\right]_0^{\ln 2} = \sinh(\ln 2) - 0$$

Step 3 $\sinh(\ln 2) = \dfrac{e^{\ln 2} - e^{-\ln 2}}{2} = \dfrac{2 - \frac{1}{2}}{2} = \dfrac{3}{4}$

$$L = \frac{3}{4}$$

Worked Example 6.2 — Surface of Revolution

Find the surface area generated when $y = \sqrt{x}$ for $1 \leq x \leq 4$ is rotated $2\pi$ about the $x$-axis.

Step 1 $y = x^{1/2}$, so $\dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$, and $1 + \left(\dfrac{dy}{dx}\right)^2 = 1 + \dfrac{1}{4x} = \dfrac{4x+1}{4x}$.

Step 2 $$S = 2\pi\int_1^4 \sqrt{x}\cdot\sqrt{\frac{4x+1}{4x}}\,dx = 2\pi\int_1^4\frac{\sqrt{4x+1}}{2}\,dx = \pi\int_1^4\sqrt{4x+1}\,dx$$

Step 3 Let $u = 4x+1$, $du = 4\,dx$: $$= \pi\int_5^{17}\frac{\sqrt{u}}{4}\,du = \frac{\pi}{4}\cdot\frac{2}{3}\left[u^{3/2}\right]_5^{17} = \frac{\pi}{6}\left(17^{3/2} - 5^{3/2}\right) = \frac{\pi}{6}\left(17\sqrt{17} - 5\sqrt{5}\right)$$

7. Second-Order Differential Equations

Second-order linear ODEs with constant coefficients have the general form: $$a\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)$$ The general solution is the sum of the complementary function (CF) and a particular integral (PI).

Complementary Function

Solve the auxiliary equation $am^2 + bm + c = 0$. The CF depends on the nature of the roots:

Particular Integral

The PI is a specific solution to the full equation. Choose a trial PI of the same form as $f(x)$:

Worked Example 7.1 — Second-Order ODE

Solve $\dfrac{d^2y}{dx^2} - 3\dfrac{dy}{dx} + 2y = 4e^{3x}$, given $y(0) = 1$ and $y'(0) = 0$.

Step 1: CF Auxiliary equation: $m^2 - 3m + 2 = (m-1)(m-2) = 0$, so $m = 1, 2$. $$y_{CF} = Ae^x + Be^{2x}$$

Step 2: PI Try $y_p = \lambda e^{3x}$. Then $y_p'' - 3y_p' + 2y_p = 9\lambda e^{3x} - 9\lambda e^{3x} + 2\lambda e^{3x} = 2\lambda e^{3x} = 4e^{3x}$, so $\lambda = 2$. $$y_p = 2e^{3x}$$

Step 3: General Solution $$y = Ae^x + Be^{2x} + 2e^{3x}$$

Step 4: Apply Conditions $y(0) = A + B + 2 = 1 \Rightarrow A + B = -1$.

$y' = Ae^x + 2Be^{2x} + 6e^{3x}$, so $y'(0) = A + 2B + 6 = 0 \Rightarrow A + 2B = -6$.

Subtracting: $B = -5$, $A = 4$. $$\boxed{y = 4e^x - 5e^{2x} + 2e^{3x}}$$

Integrating Factor for First-Order ODEs

The first-order linear ODE $\dfrac{dy}{dx} + P(x)y = Q(x)$ is solved using the integrating factor $\mu(x) = e^{\int P(x)\,dx}$:

$$\frac{d}{dx}[\mu y] = \mu Q(x) \implies \mu y = \int \mu Q(x)\,dx$$

Worked Example 7.2 — Integrating Factor

Solve $x\dfrac{dy}{dx} + 2y = x^3$, $x > 0$.

Step 1 Rewrite in standard form: $\dfrac{dy}{dx} + \dfrac{2}{x}y = x^2$.

Step 2 Integrating factor: $\mu = e^{\int (2/x)\,dx} = e^{2\ln x} = x^2$.

Step 3 Multiply through by $x^2$: $\dfrac{d}{dx}[x^2 y] = x^4$.

Step 4 Integrate: $x^2 y = \dfrac{x^5}{5} + C$, so $y = \dfrac{x^3}{5} + Cx^{-2}$.

8. The $t$-Substitution (Weierstrass Substitution)

The Weierstrass substitution $t = \tan\!\left(\dfrac{x}{2}\right)$ converts rational expressions in $\sin x$ and $\cos x$ into rational expressions in $t$, which can then be integrated by standard methods.

The Weierstrass Substitution Formulae

Let $t = \tan\!\left(\dfrac{x}{2}\right)$. Then:

$$\sin x = \frac{2t}{1+t^2}, \qquad \cos x = \frac{1-t^2}{1+t^2}, \qquad \tan x = \frac{2t}{1-t^2}, \qquad dx = \frac{2}{1+t^2}\,dt$$

Worked Example 8.1 — $t$-Substitution

Evaluate $\displaystyle\int \frac{1}{3 + 5\cos x}\,dx$.

Step 1 Substitute $\cos x = \dfrac{1-t^2}{1+t^2}$ and $dx = \dfrac{2}{1+t^2}\,dt$: $$\int\frac{1}{3 + 5\cdot\frac{1-t^2}{1+t^2}}\cdot\frac{2}{1+t^2}\,dt = \int\frac{1}{\frac{3(1+t^2)+5(1-t^2)}{1+t^2}}\cdot\frac{2}{1+t^2}\,dt$$

Step 2 Simplify the denominator: $3 + 3t^2 + 5 - 5t^2 = 8 - 2t^2$: $$= \int\frac{2}{8-2t^2}\,dt = \int\frac{1}{4-t^2}\,dt = \int\frac{1}{(2-t)(2+t)}\,dt$$

Step 3 Partial fractions: $\dfrac{1}{(2-t)(2+t)} = \dfrac{1/4}{2+t} + \dfrac{1/4}{2-t}$: $$= \frac{1}{4}\ln\!\left|\frac{2+t}{2-t}\right| + C = \frac{1}{4}\ln\!\left|\frac{2+\tan(x/2)}{2-\tan(x/2)}\right| + C$$

9. Practice Problems

Problem 1 — Polar Area

Find the area enclosed by one loop of the rose curve $r = 4\cos 2\theta$.

Show Solution

The curve $r = 4\cos 2\theta$ has 4 petals. One petal in the positive $x$-direction occurs for $-\pi/4 \leq \theta \leq \pi/4$ (where $r \geq 0$).

$$A = \frac{1}{2}\int_{-\pi/4}^{\pi/4}(4\cos 2\theta)^2\,d\theta = 8\int_0^{\pi/4}16\cos^2 2\theta\cdot\frac{1}{2}\,d\theta$$

Wait, let us be careful: $A = \frac{1}{2}\int_{-\pi/4}^{\pi/4}16\cos^2 2\theta\,d\theta = 8\int_0^{\pi/4}\cos^2 2\theta\,d\theta \cdot 2 = 16\int_0^{\pi/4}\frac{1+\cos 4\theta}{2}\,d\theta$

$= 8\left[\theta + \frac{\sin 4\theta}{4}\right]_0^{\pi/4} = 8\left(\frac{\pi}{4} + 0\right) = 2\pi$

Problem 2 — Hyperbolic Functions

Prove that $\cosh^2 x - \sinh^2 x = 1$ using the exponential definitions, and verify that $(\tanh x)' = \text{sech}^2 x$.

Show Solution

$\cosh^2 x - \sinh^2 x = \left(\dfrac{e^x+e^{-x}}{2}\right)^2 - \left(\dfrac{e^x-e^{-x}}{2}\right)^2$

$= \dfrac{(e^x+e^{-x})^2 - (e^x-e^{-x})^2}{4} = \dfrac{4e^x e^{-x}}{4} = 1$ ✓

$\dfrac{d}{dx}\tanh x = \dfrac{d}{dx}\!\left(\dfrac{\sinh x}{\cosh x}\right) = \dfrac{\cosh x \cdot \cosh x - \sinh x \cdot \sinh x}{\cosh^2 x} = \dfrac{1}{\cosh^2 x} = \text{sech}^2 x$ ✓

Problem 3 — Reduction Formula

Let $I_n = \displaystyle\int_0^1 x^n e^x\,dx$. Show that $I_n = e - nI_{n-1}$, and hence evaluate $I_3$.

Show Solution

Integrate by parts with $u = x^n$, $dv = e^x\,dx$: $I_n = [x^n e^x]_0^1 - n\int_0^1 x^{n-1}e^x\,dx = e - nI_{n-1}$ ✓

$I_0 = [e^x]_0^1 = e - 1$

$I_1 = e - I_0 = e - (e-1) = 1$

$I_2 = e - 2I_1 = e - 2$

$I_3 = e - 3I_2 = e - 3(e-2) = e - 3e + 6 = 6 - 2e$

Problem 4 — Second-Order ODE

Find the general solution of $\dfrac{d^2y}{dx^2} + 4y = \sin 2x$.

Show Solution

CF: Auxiliary equation $m^2 + 4 = 0$, so $m = \pm 2i$. Thus $y_{CF} = A\cos 2x + B\sin 2x$.

PI: Since $\sin 2x$ is in the CF, try $y_p = x(\lambda\cos 2x + \mu\sin 2x)$.

$y_p' = \lambda\cos 2x + \mu\sin 2x + x(-2\lambda\sin 2x + 2\mu\cos 2x)$

$y_p'' = -4\lambda\sin 2x + 4\mu\cos 2x + x(-4\lambda\cos 2x - 4\mu\sin 2x)$

$y_p'' + 4y_p = -4\lambda\sin 2x + 4\mu\cos 2x = \sin 2x$

So $\mu = 0$ and $\lambda = -\dfrac{1}{4}$. PI: $y_p = -\dfrac{x\cos 2x}{4}$.

$$y = A\cos 2x + B\sin 2x - \frac{x\cos 2x}{4}$$

Problem 5 — Inverse Hyperbolic Integration

Evaluate $\displaystyle\int_0^2 \frac{1}{\sqrt{x^2+9}}\,dx$, giving your answer in terms of $\ln$.

Show Solution

Using the standard result $\displaystyle\int\frac{1}{\sqrt{x^2+a^2}}\,dx = \ln(x+\sqrt{x^2+a^2}) + C$ with $a = 3$:

$$\left[\ln(x+\sqrt{x^2+9})\right]_0^2 = \ln(2+\sqrt{13}) - \ln(0+3) = \ln\!\left(\frac{2+\sqrt{13}}{3}\right)$$

Problem 6 — Arc Length (Parametric)

Find the length of the curve $x = 3t^2$, $y = 2t^3$ for $0 \leq t \leq 1$.

Show Solution

$\dfrac{dx}{dt} = 6t$, $\dfrac{dy}{dt} = 6t^2$, so $\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 = 36t^2 + 36t^4 = 36t^2(1+t^2)$.

$$L = \int_0^1\sqrt{36t^2(1+t^2)}\,dt = 6\int_0^1 t\sqrt{1+t^2}\,dt$$

Let $u = 1+t^2$, $du = 2t\,dt$: $$= 6\int_1^2\frac{\sqrt{u}}{2}\,du = 3\cdot\frac{2}{3}\left[u^{3/2}\right]_1^2 = 2(2\sqrt{2}-1) = 4\sqrt{2}-2$$

Problem 7 — $t$-Substitution

Evaluate $\displaystyle\int_0^{\pi/2}\frac{1}{1+\sin x}\,dx$ using the Weierstrass substitution.

Show Solution

Let $t = \tan(x/2)$. When $x = 0$, $t = 0$; when $x = \pi/2$, $t = 1$.

$\sin x = \dfrac{2t}{1+t^2}$, $dx = \dfrac{2}{1+t^2}\,dt$.

$$\int_0^1\frac{1}{1+\frac{2t}{1+t^2}}\cdot\frac{2}{1+t^2}\,dt = \int_0^1\frac{2}{1+t^2+2t}\,dt = \int_0^1\frac{2}{(1+t)^2}\,dt$$

$$= 2\left[-\frac{1}{1+t}\right]_0^1 = 2\left(-\frac{1}{2}+1\right) = 1$$

Problem 8 — Polar Curve Identification

Convert the Cartesian equation $x^2 + y^2 = 4x$ to polar form. Identify and describe the curve.

Show Solution

Substitute $x = r\cos\theta$ and $x^2 + y^2 = r^2$:

$r^2 = 4r\cos\theta$, so dividing by $r$ (for $r \neq 0$): $r = 4\cos\theta$.

This is a circle with centre $(2, 0)$ in Cartesian coordinates and radius 2, passing through the origin. In polar form $r = 2a\cos\theta$ with $a = 2$, this confirms it is a circle passing through the pole with diameter along the initial line.

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