A-Level Further Mathematics – Chapter 9: Further Calculus Further Pure
Contents
This chapter extends the integration techniques from A-Level Mathematics into genuinely new territory. We begin by relaxing the assumption that integrands are bounded and intervals are finite, giving us improper integrals with their subtle convergence questions. We then introduce two rich families of functions — inverse trigonometric and hyperbolic functions — that arise naturally as antiderivatives of algebraic expressions. Finally, we apply integration to compute geometric quantities (arc length and surface area) and develop reduction formulae that elegantly connect integrals with different indices.
9.1 Improper Integrals
Definition — Improper Integrals
An integral is improper if:
- One or both limits of integration are infinite, or
- The integrand has a vertical asymptote (blow-up) within or at the boundary of the interval.
We define improper integrals as limits of proper integrals:
$$\int_a^{\infty} f(x)\,dx = \lim_{R\to\infty}\int_a^{R}f(x)\,dx$$
$$\int_a^{b}f(x)\,dx = \lim_{\varepsilon\to 0^+}\int_{a+\varepsilon}^{b}f(x)\,dx \quad \text{if } f \text{ blows up at } x=a$$
If the limit exists and is finite, the integral converges; otherwise it diverges.
Theorem 9.1 — The $p$-Integral Test
$$\int_1^{\infty}\frac{1}{x^p}\,dx \text{ converges if and only if } p > 1$$
When it converges: $\displaystyle\int_1^{\infty}\frac{1}{x^p}\,dx = \frac{1}{p-1}$.
Proof. For $p\ne1$: $\int_1^R x^{-p}\,dx = \Bigl[\frac{x^{1-p}}{1-p}\Bigr]_1^R = \frac{R^{1-p}-1}{1-p}$. If $p>1$ then $1-p<0$ and $R^{1-p}\to0$, giving $\frac{1}{p-1}$. If $p<1$ then $R^{1-p}\to\infty$, diverging. For $p=1$: $\int_1^R\frac{1}{x}\,dx = \ln R\to\infty$, also divergent.
Example 9.1.1 — A convergent improper integral
Evaluate $\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx$.
$\int_1^R x^{-2}\,dx = \Bigl[-x^{-1}\Bigr]_1^R = -\frac{1}{R}+1$. As $R\to\infty$: $\int_1^{\infty}\frac{1}{x^2}\,dx = 1$.
Example 9.1.2 — A divergent improper integral
Show that $\displaystyle\int_1^{\infty}\frac{1}{\sqrt{x}}\,dx$ diverges.
$\int_1^R x^{-1/2}\,dx = \Bigl[2x^{1/2}\Bigr]_1^R = 2\sqrt{R}-2 \to \infty$ as $R\to\infty$. The integral diverges.
Example 9.1.3 — Blow-up at the lower limit
Evaluate $\displaystyle\int_0^1\frac{1}{\sqrt{x}}\,dx$.
$\int_\varepsilon^1 x^{-1/2}\,dx = \Bigl[2\sqrt{x}\Bigr]_\varepsilon^1 = 2 - 2\sqrt{\varepsilon}\to 2$ as $\varepsilon\to0^+$. The integral converges to $2$.
Example 9.1.4 — Integral of $e^{-x}$
Evaluate $\displaystyle\int_0^{\infty}e^{-x}\,dx$ and $\displaystyle\int_0^{\infty}xe^{-x}\,dx$.
$\int_0^R e^{-x}\,dx = 1-e^{-R}\to 1$.
$\int_0^R xe^{-x}\,dx = \Bigl[-xe^{-x}-e^{-x}\Bigr]_0^R = -(R+1)e^{-R}+1\to 1$ (since $Re^{-R}\to0$).
Example 9.1.5 — Deciding convergence by comparison
Without evaluating, determine whether $\displaystyle\int_1^{\infty}\frac{\sin^2 x}{x^2}\,dx$ converges.
Since $0 \le \sin^2 x \le 1$, we have $0 \le \frac{\sin^2 x}{x^2} \le \frac{1}{x^2}$. Since $\int_1^{\infty}\frac{1}{x^2}\,dx = 1$ converges, by comparison the given integral also converges.
Exam Tip
Always replace the infinite limit (or the blow-up point) with a parameter and take the limit at the end — do not simply "substitute $\infty$" into the antiderivative. Examiners specifically penalise the omission of the limit notation, even if the final answer is correct.
9.2 Inverse Trigonometric Integrals
Theorem 9.2 — Standard Inverse Trigonometric Integrals
$$\int\frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C$$
$$\int\frac{1}{1+x^2}\,dx = \arctan x + C$$
General forms with parameter $a > 0$:
$$\int\frac{1}{\sqrt{a^2-x^2}}\,dx = \arcsin\!\left(\frac{x}{a}\right) + C$$
$$\int\frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C$$
Example 9.2.1 — Basic arctan integral
Evaluate $\displaystyle\int_0^{\sqrt{3}}\frac{1}{1+x^2}\,dx$.
$\Bigl[\arctan x\Bigr]_0^{\sqrt{3}} = \arctan\sqrt{3} - \arctan 0 = \frac{\pi}{3} - 0 = \frac{\pi}{3}$.
Example 9.2.2 — General arctan form
Evaluate $\displaystyle\int_0^{2}\frac{1}{4+x^2}\,dx$.
$a=2$: $\int_0^2\frac{1}{4+x^2}\,dx = \frac{1}{2}\Bigl[\arctan\frac{x}{2}\Bigr]_0^2 = \frac{1}{2}\Bigl(\arctan 1 - 0\Bigr) = \frac{1}{2}\cdot\frac{\pi}{4} = \frac{\pi}{8}$.
Example 9.2.3 — Completing the square first
Evaluate $\displaystyle\int\frac{1}{x^2+4x+13}\,dx$.
Complete the square: $x^2+4x+13=(x+2)^2+9$.
$\int\frac{1}{(x+2)^2+9}\,dx = \frac{1}{3}\arctan\!\left(\frac{x+2}{3}\right)+C$.
Example 9.2.4 — Arcsin form
Evaluate $\displaystyle\int_0^{1}\frac{3}{\sqrt{9-x^2}}\,dx$.
$\int_0^1\frac{3}{\sqrt{9-x^2}}\,dx = 3\Bigl[\arcsin\frac{x}{3}\Bigr]_0^1 = 3\arcsin\frac{1}{3}$.
Example 9.2.5 — Linear substitution needed
Evaluate $\displaystyle\int\frac{1}{\sqrt{4-9x^2}}\,dx$.
Write as $\frac{1}{2\sqrt{1-(3x/2)^2}}$. Let $u=\tfrac{3x}{2}$, $du=\tfrac{3}{2}dx$.
$\int\frac{1}{\sqrt{4-9x^2}}\,dx = \frac{1}{3}\arcsin\!\left(\frac{3x}{2}\right)+C$.
Example 9.2.6 — Improper arctan integral
Show that $\displaystyle\int_0^{\infty}\frac{1}{1+x^2}\,dx = \frac{\pi}{2}$.
$\int_0^R\frac{1}{1+x^2}\,dx = \arctan R - 0 = \arctan R \to \frac{\pi}{2}$ as $R\to\infty$. ✓
9.3 Hyperbolic Functions
Definition — Hyperbolic Functions
$$\sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}, \qquad \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x-e^{-x}}{e^x+e^{-x}}$$
Further functions: $\text{cosech}\,x = \frac{1}{\sinh x}$, $\text{sech}\,x = \frac{1}{\cosh x}$, $\coth x = \frac{\cosh x}{\sinh x}$.
Theorem 9.3 — Fundamental Identities and Derivatives
Key identity (Osborn's rule analogue): $\cosh^2 x - \sinh^2 x = 1$.
Further identities: $1 - \tanh^2 x = \text{sech}^2 x$; $\coth^2 x - 1 = \text{cosech}^2 x$.
Derivatives:
| $f(x)$ | $f'(x)$ |
|---|---|
| $\sinh x$ | $\cosh x$ |
| $\cosh x$ | $\sinh x$ |
| $\tanh x$ | $\text{sech}^2 x$ |
Note: Unlike circular functions, $\frac{d}{dx}\cosh x = +\sinh x$ (not minus).
Double Angle and Addition Formulae for Hyperbolic Functions
By Osborn's rule (replace circular with hyperbolic and negate any product of two $\sinh$ terms):
- $\sinh(A+B) = \sinh A\cosh B + \cosh A\sinh B$
- $\cosh(A+B) = \cosh A\cosh B + \sinh A\sinh B$ (note: $+$, not $-$)
- $\sinh 2x = 2\sinh x\cosh x$
- $\cosh 2x = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1 = 1 + 2\sinh^2 x$
Example 9.3.1 — Proving the fundamental identity
Prove that $\cosh^2 x - \sinh^2 x = 1$ directly from the exponential definitions.
$\cosh^2 x - \sinh^2 x = \left(\frac{e^x+e^{-x}}{2}\right)^2 - \left(\frac{e^x-e^{-x}}{2}\right)^2$
$= \frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{4} = \frac{4e^x e^{-x}}{4} = \frac{4}{4} = 1$. ✓
Example 9.3.2 — Integrating hyperbolic functions
Find $\displaystyle\int_0^{\ln 2}\sinh x\,dx$ and $\displaystyle\int_0^1\tanh x\,dx$.
$\int_0^{\ln 2}\sinh x\,dx = \Bigl[\cosh x\Bigr]_0^{\ln 2} = \cosh(\ln 2)-1$.
$\cosh(\ln 2) = \frac{e^{\ln 2}+e^{-\ln 2}}{2} = \frac{2+\tfrac{1}{2}}{2} = \frac{5}{4}$. So the integral equals $\frac{5}{4}-1 = \frac{1}{4}$.
$\int_0^1\tanh x\,dx = \int_0^1\frac{\sinh x}{\cosh x}\,dx = \Bigl[\ln\cosh x\Bigr]_0^1 = \ln\cosh 1$.
Example 9.3.3 — Solving a hyperbolic equation
Solve $5\cosh x - 4\sinh x = 8$, giving your answer in exact logarithmic form.
Substitute definitions: $5\cdot\frac{e^x+e^{-x}}{2} - 4\cdot\frac{e^x-e^{-x}}{2} = 8$, giving $\frac{e^x+9e^{-x}}{2}=8$, so $e^x+9e^{-x}=16$.
Multiply by $e^x$: $e^{2x}-16e^x+9=0$. Let $u=e^x$: $u^2-16u+9=0$, $u=\frac{16\pm\sqrt{220}}{2}=8\pm\sqrt{55}$.
Since $u=e^x>0$, both solutions are positive: $x = \ln(8+\sqrt{55})$ or $x=\ln(8-\sqrt{55})$.
Exam Tip
Osborn's rule is a quick memory aid: to convert a circular trigonometric identity into the corresponding hyperbolic identity, replace every occurrence of a product $\sin A\sin B$ (or $\sin^2$) with $-\sinh A\sinh B$ (or $-\sinh^2$). Every other function maps directly to its hyperbolic counterpart.
9.4 Inverse Hyperbolic Functions
Theorem 9.4 — Logarithmic Forms of Inverse Hyperbolic Functions
$$\text{arcsinh}\,x = \ln\!\left(x + \sqrt{x^2+1}\right), \quad x \in \mathbb{R}$$
$$\text{arccosh}\,x = \ln\!\left(x + \sqrt{x^2-1}\right), \quad x \ge 1$$
$$\text{arctanh}\,x = \frac{1}{2}\ln\!\left(\frac{1+x}{1-x}\right), \quad |x|<1$$
Theorem 9.5 — Inverse Hyperbolic Integrals
$$\int\frac{1}{\sqrt{x^2+a^2}}\,dx = \text{arcsinh}\!\left(\frac{x}{a}\right)+C = \ln\!\left(x+\sqrt{x^2+a^2}\right)+C$$
$$\int\frac{1}{\sqrt{x^2-a^2}}\,dx = \text{arccosh}\!\left(\frac{x}{a}\right)+C = \ln\!\left(x+\sqrt{x^2-a^2}\right)+C \quad (x>a)$$
$$\int\frac{1}{a^2-x^2}\,dx = \frac{1}{a}\text{arctanh}\!\left(\frac{x}{a}\right)+C = \frac{1}{2a}\ln\!\left(\frac{a+x}{a-x}\right)+C \quad (|x|
Example 9.4.1 — Proving the arcsinh logarithmic form
Show that $\text{arcsinh}\,x = \ln(x+\sqrt{x^2+1})$.
Let $y = \text{arcsinh}\,x$, so $x = \sinh y = \frac{e^y-e^{-y}}{2}$. Then $2x = e^y - e^{-y}$, so $e^{2y}-2xe^y-1=0$.
By the quadratic formula: $e^y = \frac{2x\pm\sqrt{4x^2+4}}{2} = x\pm\sqrt{x^2+1}$. Since $e^y>0$, we take the $+$ sign: $e^y = x+\sqrt{x^2+1}$, so $y = \ln(x+\sqrt{x^2+1})$. ✓
Example 9.4.2 — Standard arcsinh integral
Evaluate $\displaystyle\int_0^{3}\frac{1}{\sqrt{x^2+9}}\,dx$, giving your answer in exact form.
$a=3$: $\Bigl[\text{arcsinh}\!\left(\tfrac{x}{3}\right)\Bigr]_0^3 = \text{arcsinh}(1)-0 = \ln(1+\sqrt{2})$.
Example 9.4.3 — Completing the square before integrating
Evaluate $\displaystyle\int\frac{1}{\sqrt{x^2-6x+10}}\,dx$.
Complete the square: $x^2-6x+10 = (x-3)^2+1$.
$\int\frac{1}{\sqrt{(x-3)^2+1}}\,dx = \text{arcsinh}(x-3)+C = \ln\!\left(x-3+\sqrt{(x-3)^2+1}\right)+C$.
Example 9.4.4 — Arccosh integral
Evaluate $\displaystyle\int_2^5\frac{1}{\sqrt{x^2-1}}\,dx$.
$a=1$: $\Bigl[\text{arccosh}\,x\Bigr]_2^5 = \ln(5+\sqrt{24})-\ln(2+\sqrt{3}) = \ln(5+2\sqrt{6})-\ln(2+\sqrt{3})$.
9.5 Arc Length and Surface Area
Theorem 9.6 — Arc Length Formulae
Cartesian: For $y=f(x)$ from $x=a$ to $x=b$:
$$L = \int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$
Parametric: For $x=x(t)$, $y=y(t)$ from $t=t_1$ to $t=t_2$:
$$L = \int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$
Theorem 9.7 — Surface Area of Revolution
When the curve $y=f(x)$, $a\le x\le b$, is rotated through $2\pi$ about the $x$-axis, the surface area generated is:
$$S = 2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$$
Example 9.5.1 — Arc length of a parabola
Find the arc length of $y = \frac{x^2}{2}$ from $x=0$ to $x=1$.
$\frac{dy}{dx} = x$. $L = \int_0^1\sqrt{1+x^2}\,dx$.
Using the standard result with $a=1$: $\int\sqrt{1+x^2}\,dx = \tfrac{x\sqrt{1+x^2}}{2}+\tfrac{1}{2}\text{arcsinh}\,x + C$.
$L = \Bigl[\tfrac{x\sqrt{1+x^2}}{2}+\tfrac{1}{2}\ln(x+\sqrt{1+x^2})\Bigr]_0^1 = \frac{\sqrt{2}}{2}+\frac{\ln(1+\sqrt{2})}{2}$.
Example 9.5.2 — Parametric arc length: the cycloid
For the cycloid $x = t - \sin t$, $y = 1 - \cos t$, find the arc length for one arch $0\le t\le 2\pi$.
$\frac{dx}{dt} = 1-\cos t$, $\frac{dy}{dt} = \sin t$.
$\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 = (1-\cos t)^2+\sin^2 t = 1-2\cos t+\cos^2 t+\sin^2 t = 2-2\cos t = 4\sin^2\!\tfrac{t}{2}$.
$L = \int_0^{2\pi}2\left|\sin\frac{t}{2}\right|\,dt = 2\int_0^{2\pi}\sin\frac{t}{2}\,dt = 2\Bigl[-2\cos\frac{t}{2}\Bigr]_0^{2\pi} = 2[2+2] = 8$.
Example 9.5.3 — Surface area of a sphere
Derive the formula $S = 4\pi r^2$ by rotating a semicircle $y = \sqrt{r^2-x^2}$ about the $x$-axis.
$\frac{dy}{dx} = \frac{-x}{\sqrt{r^2-x^2}}$. $1+\left(\frac{dy}{dx}\right)^2 = 1+\frac{x^2}{r^2-x^2} = \frac{r^2}{r^2-x^2}$.
$S = 2\pi\int_{-r}^{r}\sqrt{r^2-x^2}\cdot\frac{r}{\sqrt{r^2-x^2}}\,dx = 2\pi\int_{-r}^{r}r\,dx = 2\pi r\cdot 2r = 4\pi r^2$. ✓
Example 9.5.4 — Surface area of a cone
Find the curved surface area generated when $y = \tfrac{x}{2}$, $0\le x\le 4$, is rotated about the $x$-axis.
$\frac{dy}{dx} = \tfrac{1}{2}$. $\sqrt{1+\tfrac{1}{4}} = \frac{\sqrt{5}}{2}$.
$S = 2\pi\int_0^4\frac{x}{2}\cdot\frac{\sqrt{5}}{2}\,dx = \frac{\pi\sqrt{5}}{2}\int_0^4 x\,dx = \frac{\pi\sqrt{5}}{2}\cdot 8 = 4\pi\sqrt{5}$.
Exam Tip
Arc length and surface area integrals often simplify because $1+(dy/dx)^2$ is a perfect square. Always look for this simplification before attempting a more complicated integration technique. Check that you have the correct formula: arc length does not include the $2\pi y$ factor; surface area does.
9.6 Reduction Formulae
Definition — Reduction Formula
A reduction formula expresses $I_n = \int f_n(x)\,dx$ in terms of $I_{n-1}$ or $I_{n-2}$ (or another $I_k$ with $k < n$), allowing us to compute $I_n$ for any $n$ by repeated application.
The standard technique is integration by parts, setting up the result in terms of a lower-index integral.
Example 9.6.1 — Reduction formula for $\int \sin^n x\,dx$
Let $I_n = \int_0^{\pi/2}\sin^n x\,dx$. Derive the reduction formula $I_n = \frac{n-1}{n}I_{n-2}$.
Step 1 Write $\sin^n x = \sin^{n-1}x\cdot\sin x$ and integrate by parts: let $u=\sin^{n-1}x$, $dv=\sin x\,dx$.
$du=(n-1)\sin^{n-2}x\cos x\,dx$, $v=-\cos x$.
Step 2 $I_n = \Bigl[-\sin^{n-1}x\cos x\Bigr]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,dx$.
The boundary term vanishes at both limits. Replace $\cos^2 x = 1-\sin^2 x$:
Step 3 $I_n = (n-1)I_{n-2}-(n-1)I_n$. Hence $nI_n = (n-1)I_{n-2}$, so $\boxed{I_n = \frac{n-1}{n}I_{n-2}}$.
Example 9.6.2 — Evaluating $I_5$ and $I_6$ using the reduction formula
Find $\displaystyle\int_0^{\pi/2}\sin^5 x\,dx$ and $\displaystyle\int_0^{\pi/2}\sin^6 x\,dx$.
Base cases: $I_0 = \frac{\pi}{2}$, $I_1 = 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}$.
$I_6 = \frac{5}{6}I_4 = \frac{5}{6}\cdot\frac{3}{4}I_2 = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}I_0 = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{5\pi}{32}$.
Theorem 9.8 — Wallis's Formula
From the reduction formula $I_n = \frac{n-1}{n}I_{n-2}$ with $I_n = \int_0^{\pi/2}\sin^n x\,dx$:
$$\frac{\pi}{2} = \lim_{n\to\infty}\frac{(2n)!!}{(2n-1)!!}\cdot\frac{1}{\sqrt{n}} \quad \text{equivalently: } \frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots$$
This is Wallis's product for $\pi$, derived entirely from the reduction formula.
Example 9.6.3 — Reduction formula for $\int x^n e^x\,dx$
Let $I_n = \int x^n e^x\,dx$. Show that $I_n = x^n e^x - nI_{n-1}$.
Integrate by parts: $u = x^n$, $dv = e^x\,dx$. Then $du = nx^{n-1}\,dx$, $v = e^x$.
$I_n = x^n e^x - n\int x^{n-1}e^x\,dx = x^n e^x - nI_{n-1}$. ✓
For example: $I_2 = x^2 e^x - 2I_1 = x^2 e^x - 2(xe^x-e^x) + C = (x^2-2x+2)e^x + C$.
Example 9.6.4 — Reduction formula for $\int \sec^n x\,dx$
Let $I_n = \int\sec^n x\,dx$. By writing $\sec^n x = \sec^{n-2}x\cdot\sec^2 x$ and integrating by parts, show that $I_n = \frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}I_{n-2}$.
Let $u=\sec^{n-2}x$, $dv=\sec^2 x\,dx$. Then $du=(n-2)\sec^{n-2}x\tan x\,dx$, $v=\tan x$.
$I_n = \sec^{n-2}x\tan x - (n-2)\int\sec^{n-2}x\tan^2 x\,dx$.
Replace $\tan^2 x = \sec^2 x-1$: $(n-1)I_n = \sec^{n-2}x\tan x + (n-2)I_{n-2}$. Divide by $(n-1)$. ✓
Example 9.6.5 — Definite integral via reduction
Let $I_n = \int_0^1 x^n e^{-x}\,dx$. Show that $I_n = -e^{-1} + nI_{n-1}$ and use this to find $I_3$.
Parts: $u=x^n$, $dv=e^{-x}\,dx$, $v=-e^{-x}$.
$I_n = \Bigl[-x^n e^{-x}\Bigr]_0^1+n\int_0^1 x^{n-1}e^{-x}\,dx = -e^{-1}+nI_{n-1}$. ✓
$I_0 = 1-e^{-1}$. $I_1 = -e^{-1}+I_0 = 1-2e^{-1}$. $I_2 = -e^{-1}+2I_1 = 2-5e^{-1}$. $I_3 = -e^{-1}+3I_2 = 6-16e^{-1}$.
Example 9.6.6 — Reduction for $\int \cos^n x\,dx$
Show that the same reduction formula applies: $\int_0^{\pi/2}\cos^n x\,dx = \frac{n-1}{n}\int_0^{\pi/2}\cos^{n-2}x\,dx$. Hence find $\int_0^{\pi/2}\cos^4 x\,dx$.
By symmetry (substitute $x\mapsto\frac{\pi}{2}-x$), $\int_0^{\pi/2}\cos^n x\,dx = \int_0^{\pi/2}\sin^n x\,dx = I_n$. The same reduction formula holds.
$\int_0^{\pi/2}\cos^4 x\,dx = I_4 = \frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{3\pi}{16}$.
Exam Tip
When deriving a reduction formula, the key step is recognising which function to take as $u$ and which as $dv$ in integration by parts. For $\int\sin^n x\,dx$, always split off one power of $\sin x$ for $dv$. After integration by parts, use the Pythagorean identity to replace $\cos^2 x$ (or $\sin^2 x$), then collect the $I_n$ terms on the left-hand side. Show all these steps clearly — they are worth method marks.
Practice Problems
Problem 1
Determine whether each of the following integrals converges or diverges. For convergent integrals, find the exact value.
(a) $\displaystyle\int_1^{\infty}\frac{1}{x^3}\,dx$ (b) $\displaystyle\int_0^{\infty}e^{-3x}\,dx$ (c) $\displaystyle\int_1^{\infty}\frac{1}{\sqrt[3]{x}}\,dx$
Show solution
(b) $\int_0^R e^{-3x}\,dx = \frac{1}{3}(1-e^{-3R})\to\frac{1}{3}$. Converges to $\frac{1}{3}$.
(c) $p=\frac{1}{3}<1$, diverges.
Problem 2
Evaluate $\displaystyle\int_0^{\infty}\frac{x}{(1+x^2)^2}\,dx$.
Show solution
$\frac{1}{2}\int_1^{\infty}u^{-2}\,du = \frac{1}{2}\Bigl[-u^{-1}\Bigr]_1^{\infty} = \frac{1}{2}(0+1) = \frac{1}{2}$.
Problem 3
Find $\displaystyle\int\frac{5}{x^2+2x+10}\,dx$.
Show solution
$5\cdot\frac{1}{3}\arctan\frac{x+1}{3}+C = \frac{5}{3}\arctan\frac{x+1}{3}+C$.
Problem 4
Prove from first principles that $\dfrac{d}{dx}\text{arcsinh}\,x = \dfrac{1}{\sqrt{x^2+1}}$.
Show solution
Using $\cosh^2 y-\sinh^2 y=1$: $\cosh y=\sqrt{1+\sinh^2 y}=\sqrt{1+x^2}$ (taking positive root since $\cosh>0$).
Therefore $\frac{d}{dx}\text{arcsinh}\,x=\frac{1}{\sqrt{x^2+1}}$. ✓
Problem 5
Evaluate $\displaystyle\int_0^2\frac{1}{\sqrt{x^2+4}}\,dx$, giving your answer in exact logarithmic form.
Show solution
Problem 6
Find the arc length of the curve $y = \cosh x$ from $x = 0$ to $x = 1$.
Show solution
$L=\int_0^1\cosh x\,dx=[\sinh x]_0^1=\sinh 1=\frac{e-e^{-1}}{2}$.
Problem 7
Show that $\cosh 2x = 1 + 2\sinh^2 x$ and use this to evaluate $\displaystyle\int_0^{\ln 3}\sinh^2 x\,dx$.
Show solution
$\int_0^{\ln 3}\sinh^2 x\,dx=\int_0^{\ln 3}\frac{\cosh 2x-1}{2}\,dx=\frac{1}{2}\Bigl[\frac{\sinh 2x}{2}-x\Bigr]_0^{\ln 3}$.
$\sinh(2\ln 3)=\sinh\ln 9=\frac{9-\frac{1}{9}}{2}=\frac{80}{18}=\frac{40}{9}$.
$=\frac{1}{2}\Bigl(\frac{40}{18}-\ln 3\Bigr)=\frac{20}{9}-\frac{\ln 3}{2}$.
Problem 8
Find the surface area of revolution when $y = \sqrt{x}$, $1\le x\le 4$, is rotated through $2\pi$ about the $x$-axis.
Show solution
$S=2\pi\int_1^4\sqrt{x}\cdot\frac{\sqrt{4x+1}}{2\sqrt{x}}\,dx=\pi\int_1^4\sqrt{4x+1}\,dx$.
$=\pi\Bigl[\frac{(4x+1)^{3/2}}{6}\Bigr]_1^4=\frac{\pi}{6}(17^{3/2}-5^{3/2})=\frac{\pi}{6}(17\sqrt{17}-5\sqrt{5})$.
Problem 9
Let $I_n = \int_0^{\pi/4}\tan^n x\,dx$ for $n\ge 2$. Show that $I_n + I_{n-2} = \dfrac{1}{n-1}$, and hence find $I_4$.
Show solution
$=\Bigl[\frac{\tan^{n-1}x}{n-1}\Bigr]_0^{\pi/4}=\frac{1}{n-1}$. ✓
$I_2=\int_0^{\pi/4}\tan^2 x\,dx=[\tan x-x]_0^{\pi/4}=1-\frac{\pi}{4}$.
$I_4+I_2=\frac{1}{3}$, so $I_4=\frac{1}{3}-1+\frac{\pi}{4}=\frac{\pi}{4}-\frac{2}{3}$.
Problem 10
Let $I_n = \int_0^1 x^n\sqrt{1-x}\,dx$. Show that $I_n = \frac{2n}{2n+3}I_{n-1}$ and hence find $I_3$ in exact form.
Show solution
$I_n=\Bigl[-\frac{2x^n(1-x)^{3/2}}{3}\Bigr]_0^1+\frac{2n}{3}\int_0^1 x^{n-1}(1-x)^{3/2}\,dx$.
Boundary term is $0$. Now use the same technique on $\int x^{n-1}(1-x)^{3/2}$: write $(1-x)^{3/2}=(1-x)\sqrt{1-x}$ and expand:
$\int_0^1 x^{n-1}(1-x)^{3/2}\,dx=\int_0^1 x^{n-1}(1-x)\sqrt{1-x}\,dx=I_{n-1}-I_n$... actually use the Beta function observation or simply note:
$I_n=\frac{2n}{3}(I_{n-1}-I_n)$, giving $I_n(1+\frac{2n}{3})=\frac{2n}{3}I_{n-1}$, so $I_n=\frac{2n}{2n+3}I_{n-1}$. ✓
$I_0=\int_0^1\sqrt{1-x}\,dx=\frac{2}{3}$. $I_1=\frac{2}{5}I_0=\frac{4}{15}$. $I_2=\frac{4}{7}I_1=\frac{16}{105}$. $I_3=\frac{6}{9}I_2=\frac{2}{3}\cdot\frac{16}{105}=\frac{32}{315}$.