Paper 2 · Further Pure 2 ✦

Integration Techniques

Recognise the standard forms, substitute your way through the rest, tame a whole family of integrals with one reduction formula — then measure the length of a curve and the skin of a solid.

Syllabus 2.4Standard forms on MF19Arc length & surface: memorise

What the examiner expects you to do

Integrate hyperbolic functions, recognise the standard forms $\frac{1}{\sqrt{a^{2} - x^{2}}}$ , $\frac{1}{\sqrt{a^{2} + x^{2}}}$ , $\frac{1}{\sqrt{x^{2} - a^{2}}}$ , and integrate related functions using trigonometric or hyperbolic substitutions — including completing the square first.
Derive and use reduction formulae for definite integrals.
Approximate the area under a curve by rectangles, and use them to set bounds, derive inequalities, or find limits of sums.
Use integration to find arc lengths (Cartesian, parametric or polar) and surface areas of revolution about an axis (Cartesian or parametric).

Scope note: surface areas of revolution for curves given in polar form are explicitly not required. Arc length in polar form is.

§1 Hyperbolic integrals and the standard forms

◆ Straight from the derivatives MF19

\int \sinh x d x = \cosh x + c, \int \cosh x d x = \sinh x + c, \int {sech}^{2} x d x = \tanh x + c .

One more is worth knowing because it is not listed: since $\tanh x = \frac{\sinh x}{\cosh x}$ has the derivative of its denominator on top,

\int \tanh x d x = \ln (\cosh x) + c .

No modulus signs are needed — $\cosh x > 0$ for every $x$ .

◆ The four forms to recognise on sight

Integrand	Integral	On MF19?
$\frac{1}{\sqrt{a^{2} - x^{2}}}$	$\sin^{- 1} \frac{x}{a} + c (\| x \| < a)$	given
$\frac{1}{\sqrt{a^{2} + x^{2}}}$	$\sinh^{- 1} \frac{x}{a} + c$	given
$\frac{1}{\sqrt{x^{2} - a^{2}}}$	$\cosh^{- 1} \frac{x}{a} + c (x > a)$	given
$\frac{1}{a^{2} + x^{2}}$	$\frac{1}{a} \tan^{- 1} \frac{x}{a} + c$	memorise

The tell is the square root. With a root, you get an inverse sine or an inverse hyperbolic and no $\frac{1}{a}$ outside; without a root, you get $\tan^{- 1}$ and there is a $\frac{1}{a}$ outside. Mixing those two up is the single most common error in this section.

✎ Worked example — completing the square first

Find $\int \frac{1}{\sqrt{x^{2} + 4 x + 13}} d x$ .

The quadratic is not one of the standard shapes, so complete the square: $x^{2} + 4 x + 13 = (x + 2)^{2} + 9$ . Now it is the $\sqrt{a^{2} + u^{2}}$ form with $u = x + 2$ and $a = 3$ :

\int \frac{1}{\sqrt{(x + 2)^{2} + 3^{2}}} d x = \sinh^{- 1} (\frac{x + 2}{3}) + c .

Check by differentiating: $\frac{d}{d x} \sinh^{- 1} \frac{x + 2}{3} = \frac{1 / 3}{\sqrt{1 + \frac{(x + 2)^{2}}{9}}} = \frac{1}{\sqrt{9 + (x + 2)^{2}}}$ ✓

Compare with $\int \frac{1}{x^{2} + 6 x + 13} d x = \int \frac{d x}{(x + 3)^{2} + 2^{2}} = \frac{1}{2} \tan^{- 1} (\frac{x + 3}{2}) + c$ — same idea, no root, so $\tan^{- 1}$ and a $\frac{1}{2}$ appears.

§2 Trigonometric and hyperbolic substitutions

◆ Choosing the substitution — the identity does the work

Expression	Substitute	Because it becomes
$\sqrt{a^{2} - x^{2}}$	$x = a \sin θ$	$a \cos θ$ , by $1 - \sin^{2} = \cos^{2}$
$\sqrt{a^{2} + x^{2}}$	$x = a \sinh u$	$a \cosh u$ , by $1 + \sinh^{2} = \cosh^{2}$
$\sqrt{x^{2} - a^{2}}$	$x = a \cosh u$	$a \sinh u$ , by $\cosh^{2} - 1 = \sinh^{2}$

Every one is chosen so that the identity kills the square root. Remember to convert $d x$ as well, and — for a definite integral — to change the limits rather than converting back.

✎ Worked example — a trigonometric substitution

Find $\int \sqrt{4 - x^{2}} d x$ . Put $x = 2 \sin θ$ , so $d x = 2 \cos θ d θ$ and $\sqrt{4 - x^{2}} = 2 \cos θ$ :

\int 2 \cos θ \cdot 2 \cos θ d θ = 4 \int \cos^{2} θ d θ = 2 \int (1 + \cos 2 θ) d θ = 2 θ + \sin 2 θ + c .

Convert back with $\sin θ = \frac{x}{2}$ , $\cos θ = \frac{\sqrt{4 - x^{2}}}{2}$ , and $\sin 2 θ = 2 \sin θ \cos θ$ :

= 2 \sin^{- 1} \frac{x}{2} + \frac{x \sqrt{4 - x^{2}}}{2} + c .

Check by differentiating: the two pieces give $\frac{2}{\sqrt{4 - x^{2}}}$ and $\frac{4 - 2 x^{2}}{2 \sqrt{4 - x^{2}}}$ , which add to $\frac{8 - 2 x^{2}}{2 \sqrt{4 - x^{2}}} = \sqrt{4 - x^{2}}$ ✓

✎ Worked example — a hyperbolic substitution

Find $\int \sqrt{x^{2} + 9} d x$ . Put $x = 3 \sinh u$ , so $d x = 3 \cosh u d u$ and $\sqrt{x^{2} + 9} = 3 \cosh u$ :

\int 9 \cosh^{2} u d u = \frac{9}{2} \int (1 + \cosh 2 u) d u = \frac{9}{2} u + \frac{9}{4} \sinh 2 u + c = \frac{9}{2} u + \frac{9}{2} \sinh u \cosh u + c,

using $\cosh^{2} u = \frac{1}{2} (1 + \cosh 2 u)$ and then $\sinh 2 u = 2 \sinh u \cosh u$ . With $\sinh u = \frac{x}{3}$ and $\cosh u = \frac{\sqrt{x^{2} + 9}}{3}$ ,

= \frac{9}{2} \sinh^{- 1} \frac{x}{3} + \frac{x \sqrt{x^{2} + 9}}{2} + c .

The two worked examples are the same problem with one sign changed — and the whole method changes from circular to hyperbolic to match.

★ A* tip — change the limits, do not change back

For a definite integral, convert the limits at the moment you substitute. If $x = 2 \sin θ$ and $x$ runs from $0$ to $2$ , then $θ$ runs from $0$ to $\frac{π}{2}$ — and you never have to untangle $\sin^{- 1}$ at the end. It is faster and much safer.

§3 Reduction formulae

◆ The idea

Write $I_{n}$ for an integral depending on a positive integer $n$ . A reduction formula expresses $I_{n}$ in terms of $I_{n - 1}$ or $I_{n - 2}$ , so that repeated use walks you down to a base case you can evaluate directly. The engine is nearly always integration by parts, with the integrand split so that one factor differentiates towards the lower index.

✎ Worked example — the classic

I_{n} = \int_{0}^{π / 2} \sin^{n} x d x

Split off a single $\sin x$ to be integrated, leaving $\sin^{n - 1} x$ to be differentiated:

I_{n} = \int_{0}^{π / 2} \underset{u}{\underset{⏟}{\sin^{n - 1} x}} \underset{d v}{\underset{⏟}{\sin x d x}} = [- \sin^{n - 1} x \cos x]_{0}^{π / 2} + (n - 1) \int_{0}^{π / 2} \sin^{n - 2} x \cos^{2} x d x .

The bracket vanishes at both ends: at $x = \frac{π}{2}$ because $\cos \frac{π}{2} = 0$ , and at $x = 0$ because $\sin 0 = 0$ (for $n \geq 2$ ). Now replace $\cos^{2} x$ by $1 - \sin^{2} x$ :

I_{n} = (n - 1) \int_{0}^{π / 2} \sin^{n - 2} x (1 - \sin^{2} x) d x = (n - 1) (I_{n - 2} - I_{n}) .

$I_{n}$ has appeared on both sides — collect it, which is the step that makes the whole method work:

n I_{n} = (n - 1) I_{n - 2} ⟹ I_{n} = \frac{n - 1}{n} I_{n - 2}

Base cases: $I_{0} = \int_{0}^{π / 2} 1 d x = \frac{π}{2}$ and $I_{1} = \int_{0}^{π / 2} \sin x d x = 1$ . So

I_{2} = \frac{1}{2} \cdot \frac{π}{2} = \frac{π}{4}, I_{3} = \frac{2}{3} \cdot 1 = \frac{2}{3}, I_{4} = \frac{3}{4} \cdot \frac{π}{4} = \frac{3 π}{16}, I_{5} = \frac{4}{5} \cdot \frac{2}{3} = \frac{8}{15} .

Notice the pattern: even $n$ keeps a $π$ , odd $n$ does not — a quick check that you started from the right base case.

✎ Worked example — the "consider

\frac{d}{d x} (\dots)

" style

Let $I_{n} = \int_{0}^{π / 4} \sec^{n} x d x$ . By considering $\frac{d}{d x} (\tan x \sec^{n - 2} x)$ , find a reduction formula.

Differentiate the given product, then convert $\tan^{2} x$ to $\sec^{2} x - 1$ so that only secants remain:

\frac{d}{d x} (\tan x \sec^{n - 2} x) = \sec^{n} x + (n - 2) \tan^{2} x \sec^{n - 2} x = \sec^{n} x + (n - 2) (\sec^{2} x - 1) \sec^{n - 2} x,

= (n - 1) \sec^{n} x - (n - 2) \sec^{n - 2} x .

Integrate both sides from $0$ to $\frac{π}{4}$ . On the left, the integral of a derivative is just the bracket, and $\tan \frac{π}{4} = 1$ , $\sec \frac{π}{4} = \sqrt{2}$ , $\tan 0 = 0$ :

(\sqrt{2})^{n - 2} = (n - 1) I_{n} - (n - 2) I_{n - 2} ⟹ I_{n} = \frac{2^{(n - 2) / 2} + (n - 2) I_{n - 2}}{n - 1} .

Check at $n = 2$ : the formula gives $I_{2} = \frac{2^{0} + 0}{1} = 1$ , and directly $\int_{0}^{π / 4} \sec^{2} x d x = [\tan x]_{0}^{π / 4} = 1$ ✓

When a question hands you a function to differentiate, that is the whole method — differentiate it, tidy with an identity, and integrate both sides.

★ A* tip — state the base case, and mind the parity

A reduction formula alone earns nothing if you cannot finish. Always evaluate $I_{0}$ and $I_{1}$ explicitly, then say which one your chain lands on: $I_{n}$ with $n$ even walks down to $I_{0}$ , and $n$ odd to $I_{1}$ .

§4 Rectangles: bounds and limits for sums

◆ The picture is the proof

For a decreasing positive function $f$ on $[r, r + 1]$ , the rectangle of height $f (r)$ is too big and the one of height $f (r + 1)$ is too small:

f (r + 1) < \int_{r}^{r + 1} f (x) d x < f (r) .

Adding these over a range of $r$ turns a sum into an integral sandwich — which is exactly how you set bounds on a sum you cannot evaluate.

✎ Worked example — bounding the harmonic sum

Show that $\ln (n + 1) < \sum_{r = 1}^{n} \frac{1}{r} \leq 1 + \ln n$ .

Lower bound. Take $f (x) = \frac{1}{x}$ , which is decreasing. On $[r, r + 1]$ we have $\frac{1}{x} < \frac{1}{r}$ , so $\int_{r}^{r + 1} \frac{d x}{x} < \frac{1}{r}$ . Summing for $r = 1, \dots, n$ , the integrals join end to end:

\int_{1}^{n + 1} \frac{d x}{x} < \sum_{r = 1}^{n} \frac{1}{r} ⟹ \ln (n + 1) < \sum_{r = 1}^{n} \frac{1}{r} .

Upper bound. On $[r - 1, r]$ we have $\frac{1}{x} > \frac{1}{r}$ , so $\frac{1}{r} < \int_{r - 1}^{r} \frac{d x}{x}$ . Summing for $r = 2, \dots, n$ and adding the first term separately:

\sum_{r = 1}^{n} \frac{1}{r} = 1 + \sum_{r = 2}^{n} \frac{1}{r} < 1 + \int_{1}^{n} \frac{d x}{x} = 1 + \ln n (n \geq 2),

with equality when $n = 1$ , where both sides are $1$ . Check at $n = 4$ : $\ln 5 = 1.609 < \frac{25}{12} = 2.083 \leq 1 + \ln 4 = 2.386$ ✓

◆ Rectangles whose width tends to zero — a sum becomes an integral

Slicing $[0,1]$ into $n$ strips of width $\frac{1}{n}$ and taking the height at the right-hand end of each strip,

\lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} f (\frac{r}{n}) = \int_{0}^{1} f (x) d x .

So a limit of a sum can be read off as an integral. For instance

\lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n + r} = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + \frac{r}{n}} = \int_{0}^{1} \frac{d x}{1 + x} = \ln 2 .

The move that unlocks these is forcing a $\frac{1}{n}$ out at the front so that what remains is a function of $\frac{r}{n}$ alone.

§5 Arc length and surface area of revolution

◆ The formulas none of these are on MF19

Arc length

s = \int \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x, s = \int \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t, s = \int \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ .

Surface area of revolution — multiply the arc-length element by $2 π \times (distance to the axis)$ :

S_{x -axis} = 2 π \int y d s, S_{y -axis} = 2 π \int x d s,

where $d s$ is whichever arc-length element above matches the way the curve is given. Surface areas from polar equations are not required.

Where

d s

comes from

A short piece of curve is almost a straight line, so by Pythagoras $δ s \approx \sqrt{δ x^{2} + δ y^{2}}$ . Factor out whichever variable you are integrating with respect to:

δ s \approx \sqrt{1 + {(\frac{δ y}{δ x})}^{2}} δ x or δ s \approx \sqrt{{(\frac{δ x}{δ t})}^{2} + {(\frac{δ y}{δ t})}^{2}} δ t,

and let the increments tend to zero. For the polar version, the element has a piece $r δ θ$ along the arc and a piece $δ r$ outwards, and these are perpendicular — so $δ s \approx \sqrt{r^{2} + {(\frac{δ r}{δ θ})}^{2}} δ θ$ . $■$

Rotating that element about the $x$ -axis sweeps a thin band of radius $y$ and width $d s$ , whose area is $2 π y d s$ — hence the surface formula.

✎ Worked example — the arc length that comes out perfectly

Find the length of $y = \cosh x$ from $x = 0$ to $x = 1$ .

$\frac{d y}{d x} = \sinh x$ , and here the identity does something remarkable:

1 + \sinh^{2} x = \cosh^{2} x ⟹ \sqrt{1 + {(\frac{d y}{d x})}^{2}} = \cosh x .

The square root disappears completely, so

s = \int_{0}^{1} \cosh x d x = [\sinh x]_{0}^{1} = \sinh 1 \approx 1.1752 .

Whenever $\cosh$ appears in an arc-length question, expect this collapse — it is why the catenary is the standard example.

✎ Worked example — a surface you already know the answer to

The arc of $y = \sqrt{a^{2} - x^{2}}$ from $x = - a$ to $x = a$ is rotated about the $x$ -axis. Find the surface area.

$\frac{d y}{d x} = - \frac{x}{y}$ , so

1 + {(\frac{d y}{d x})}^{2} = 1 + \frac{x^{2}}{y^{2}} = \frac{y^{2} + x^{2}}{y^{2}} = \frac{a^{2}}{y^{2}} ⟹ \sqrt{\cdot} = \frac{a}{y} .

The $y$ then cancels against the $y$ in the surface formula — the whole integral collapses to a constant:

S = 2 π \int_{- a}^{a} y \cdot \frac{a}{y} d x = 2 π a \int_{- a}^{a} d x = 2 π a (2 a) = 4 π a^{2} .

Which is the surface area of a sphere ✓ — a complete check on the method, and a result you already trust.

✎ Worked example — arc length in polar form

Find the total perimeter of the cardioid $r = a (1 + \cos θ)$ .

$\frac{d r}{d θ} = - a \sin θ$ , so

r^{2} + {(\frac{d r}{d θ})}^{2} = a^{2} [(1 + \cos θ)^{2} + \sin^{2} θ] = a^{2} [2 + 2 \cos θ] = 4 a^{2} \cos^{2} \frac{θ}{2},

using $1 + \cos θ = 2 \cos^{2} \frac{θ}{2}$ . Taking the square root — and noting $\cos \frac{θ}{2} \geq 0$ for $0 \leq θ \leq π$ — the curve is symmetric about the initial line, so integrate over half and double:

s = 2 \int_{0}^{π} 2 a \cos \frac{θ}{2} d θ = 2 [4 a \sin \frac{θ}{2}]_{0}^{π} = 8 a .

A curve built entirely from $π$ s and cosines has a perimeter of exactly $8 a$ — no $π$ at all.

§6 Interactive: check the calculus against the numbers

Each scenario computes the answer two ways — the exact expression from the notes, and a straight numerical integration — and compares them. Where the topic is about bounds rather than an exact value, it checks the inequality instead.

Parameter 1.0

exact vs numerical ✎

§7 Examiner traps & A* checklist

⚠ The nine most common mark-losers

Confusing $\frac{1}{\sqrt{a^{2} + x^{2}}}$ (gives $\sinh^{- 1}$ , no $\frac{1}{a}$ ) with $\frac{1}{a^{2} + x^{2}}$ (gives $\frac{1}{a} \tan^{- 1}$ ).
Not completing the square when the quadratic has an $x$ term. $x^{2} + 4 x + 13$ is not a standard form until you write it as $(x + 2)^{2} + 9$ .
Forgetting to convert $d x$ when substituting, or forgetting to change the limits on a definite integral.
In a reduction formula, failing to collect the $I_{n}$ that appears on the right-hand side.
Not justifying that the boundary bracket in the by-parts step vanishes — that is an explicit mark.
Giving a reduction formula without evaluating the base case $I_{0}$ or $I_{1}$ .
Writing the surface area as $π \int y^{2} d x$ — that is the volume of revolution. The surface is $2 π \int y d s$ .
Dropping the $d s$ factor entirely and integrating $2 π y d x$ .
Attempting a surface area from a polar equation — it is outside the syllabus, so the question will never ask for it, and if you think it has, re-read it.

★ Before you turn the page

Did I differentiate my answer back to check it against the integrand?
Is my arc length longer than the straight-line distance between the endpoints? (It must be.)
For a reduction formula, does $n = 2$ or $n = 3$ reproduce an integral I can do directly?

§8 Video explainers (curated)

▶Standard forms & substitutions
Curated search — completing the square, trig and hyperbolic substitutions.↗ ▶Reduction formulae
Curated search — deriving them by parts and finishing the chain.↗ ▶Arc length & surface area
Curated search — Cartesian, parametric and polar arc length.↗

§9 Check yourself

Score 0 / 5

Q1 · standard form

$\int \frac{1}{\sqrt{x^{2} - 9}} d x$ (for $x > 3$ ) equals:

Solution

The form $\frac{1}{\sqrt{x^{2} - a^{2}}}$ integrates to $\cosh^{- 1} \frac{x}{a}$ — all three root forms are printed on MF19. The distractors are $\frac{1}{\sqrt{a^{2} + x^{2}}}$ , $\frac{1}{\sqrt{a^{2} - x^{2}}}$ and $\frac{1}{a^{2} + x^{2}}$ . Note only the last, with no root, carries a $\frac{1}{a}$ outside.

Q2 · choosing a substitution

To find $\int \sqrt{x^{2} - 16} d x$ , the natural substitution is:

Solution

$x = 4 \cosh u$ turns $x^{2} - 16$ into $16 (\cosh^{2} u - 1) = 16 \sinh^{2} u$ , so the root becomes $4 \sinh u$ and vanishes. $x = 4 \sinh u$ suits $\sqrt{x^{2} + 16}$ , and $x = 4 \sin θ$ suits $\sqrt{16 - x^{2}}$ — each substitution is chosen to match the identity that kills the root.

Q3 · reduction formula

With $I_{n} = \int_{0}^{π / 2} \sin^{n} x d x$ and $I_{n} = \frac{n - 1}{n} I_{n - 2}$ , the value of $I_{4}$ is:

Solution

$n = 4$ is even, so the chain lands on $I_{0} = \frac{π}{2}$ : $I_{2} = \frac{1}{2} \cdot \frac{π}{2} = \frac{π}{4}$ , then $I_{4} = \frac{3}{4} \cdot \frac{π}{4} = \frac{3 π}{16}$ . The option $\frac{8}{15}$ is $I_{5}$ — odd $n$ descends to $I_{1} = 1$ and can never contain $π$ .

Q4 · surface of revolution

The arc of $y = f (x)$ between $x = a$ and $x = b$ is rotated about the $x$ -axis. The surface area generated is:

Solution

Each band has radius $y$ and width $d s$ , so its area is $2 π y d s$ . The second option is the volume of revolution, the fourth is the arc length, and the third forgets the $d s$ factor altogether. Knowing which of the three is being asked for is most of the mark.

Q5 · full method — try it on paper first

The arc of the curve $y = \cosh x$ from $x = 0$ to $x = 1$ is rotated completely about the $x$ -axis.

(i) Show that the arc length is $\sinh 1$ . (ii) Find the exact area of the surface generated.

Solution

(i) $\frac{d y}{d x} = \sinh x$ , so by $\cosh^{2} x - \sinh^{2} x \equiv 1$ ,

\sqrt{1 + {(\frac{d y}{d x})}^{2}} = \sqrt{1 + \sinh^{2} x} = \sqrt{\cosh^{2} x} = \cosh x,

taking the positive root since $\cosh x > 0$ . Hence

s = \int_{0}^{1} \cosh x d x = [\sinh x]_{0}^{1} = \sinh 1 - 0 = \sinh 1 . ■

(ii) The surface area is $2 π \int y d s$ , and part (i) already gives $d s = \cosh x d x$ :

S = 2 π \int_{0}^{1} \cosh x \cdot \cosh x d x = 2 π \int_{0}^{1} \cosh^{2} x d x .

Use the double-angle identity $\cosh^{2} x = \frac{1}{2} (1 + \cosh 2 x)$ — you cannot integrate $\cosh^{2} x$ directly:

S = 2 π \cdot \frac{1}{2} \int_{0}^{1} (1 + \cosh 2 x) d x = π [x + \frac{1}{2} \sinh 2 x]_{0}^{1} = π (1 + \frac{1}{2} \sinh 2) .

So $S = π + \frac{π}{2} \sinh 2$ .

Check: numerically $\sinh 2 = 3.6269$ , giving $S \approx 3.1416 + 5.6965 = 8.838$ . The arc is about $1.175$ long at an average height of roughly $1.2$ , so a rough band area of $2 π (1.2) (1.175) \approx 8.9$ — the right size. ✓ $■$

FP2 · Topic 11 · Integration