Further Pure — Calculus: A-level Further Maths OCR revision notes

4.08

Inverse trigonometric functions

Inverse trigonometric functions and their derivatives 4.08

Definitions

$arcsin x$ : the angle in $[- \frac{π}{2}, \frac{π}{2}]$ whose sine is $x$ , defined for $- 1 \leq x \leq 1$ .
$arccos x$ : the angle in $[0, π]$ whose cosine is $x$ , defined for $- 1 \leq x \leq 1$ .
$arctan x$ : the angle in $(- \frac{π}{2}, \frac{π}{2})$ whose tangent is $x$ , defined for all real $x$ .

Key results

$\frac{d}{d x} arcsin x = \frac{1}{1 - x ^{2}}$ , $\frac{d}{d x} arccos x = - \frac{1}{1 - x ^{2}}$ , $\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}$ .
With the chain rule: $\frac{d}{d x} arcsin (\frac{x}{a}) = \frac{1}{a ^{2} - x ^{2}}$ and $\frac{d}{d x} arctan (\frac{x}{a}) = \frac{a}{a ^{2} + x ^{2}}$ .

Method

Derivation for $arcsin$ : let $y = arcsin x$ , so $sin y = x$ . Differentiate implicitly: $cos y \frac{d y}{d x} = 1$ . Since $- \frac{π}{2} \leq y \leq \frac{π}{2}$ , $cos y \geq 0$ , so $cos y = 1 - x^{2}$ and $\frac{d y}{d x} = \frac{1}{1 - x ^{2}}$ .

In practiceShow that $\frac{d}{d x} arctan x = \frac{1}{1 + x ^{2}}$ , and differentiate $arctan 3 x$ .

$y = arctan x ⟹ tan y = x$ turn the inverse function round
$sec^{2} y \frac{d y}{d x} = 1$ differentiate implicitly
$\frac{d y}{d x} = \frac{1}{sec ^{2} y} = \frac{1}{1 + tan ^{2} y} = \frac{1}{1 + x ^{2}}$ $sec^{2} y = 1 + tan^{2} y$ ; no sign to choose
$\frac{d}{d x} arctan 3 x = \frac{3}{1 + 9 x ^{2}}$ chain rule

Notes

The restricted ranges are what make these functions well defined, and they also fix the sign of the square root in the derivation.
$arcsin x + arccos x = \frac{π}{2}$ , which is why their derivatives differ only in sign.

Integrals that give inverse trigonometric functions 4.08

Key results

$\int \frac{1}{a ^{2} + x ^{2}} d x = \frac{1}{a} arctan (\frac{x}{a}) + c$ .
$\int \frac{1}{a ^{2} - x ^{2}} d x = arcsin (\frac{x}{a}) + c$ , for $∣ x ∣ < a$ .

Method

Complete the square to reach a standard form: $\int \frac{d x}{x ^{2} + 4 x + 13} = \int \frac{d x}{( x + 2 ) ^{2} + 9} = \frac{1}{3} arctan (\frac{x + 2}{3}) + c$ .
Split a numerator so one part is a multiple of the derivative of the denominator (giving a logarithm) and the rest is a constant (giving an arctan): $\frac{2 x + 5}{x ^{2} + 1} = \frac{2 x}{x ^{2} + 1} + \frac{5}{x ^{2} + 1}$ .

In practiceFind $\int \frac{x + 4}{x ^{2} + 2 x + 5} d x$ .

$x + 4 = \frac{1}{2} (2 x + 2) + 3$ a multiple of the denominator's derivative, plus a constant
$\int \frac{x + 4}{x ^{2} + 2 x + 5} d x = \frac{1}{2} \int \frac{2 x + 2}{x ^{2} + 2 x + 5} d x + 3 \int \frac{d x}{( x + 1 ) ^{2} + 2 ^{2}}$ complete the square in the second
$= \frac{1}{2} ln (x^{2} + 2 x + 5) + \frac{3}{2} arctan (\frac{x + 1}{2}) + c$ no modulus needed: $x^{2} + 2 x + 5 > 0$

Notes

Note the factor $\frac{1}{a}$ in the arctan integral but not in the arcsin one; it comes from the chain rule above.
These standard integrals can also be derived by the substitutions $x = a tan θ$ and $x = a sin θ$ .

Worked example

Evaluate exactly:

\int_{0}^{3} \frac{d x}{9 + x ^{2}}

and:

\int_{0}^{1} \frac{d x}{4 - x ^{2}}

Show worked solution

\int_{0}^{3} \frac{d x}{9 + x ^{2}} = [\frac{1}{3} arctan \frac{x}{3}]_{0}^{3} = \frac{1}{3} \cdot \frac{π}{4} = \frac{π}{12}

And:

\int_{0}^{1} \frac{d x}{4 - x ^{2}} = [arcsin \frac{x}{2}]_{0}^{1} = arcsin \frac{1}{2} = \frac{π}{6}

4.05

Partial fractions

Partial fractions with a quadratic factor 4.05

Key results

A factor $(x^{2} + a^{2})$ in the denominator, which does not factorise over the reals, takes a linear numerator: $\frac{p x ^{2} + q x + r}{( x + b ) ( x ^{2} + a ^{2} )} = \frac{A}{x + b} + \frac{B x + C}{x ^{2} + a ^{2}}$ .

Method

Multiply through by the denominator, substitute $x = - b$ to find $A$ , then compare coefficients of $x^{2}$ and the constant term to find $B$ and $C$ .
Integrate $\frac{B x}{x ^{2} + a ^{2}}$ as $\frac{B}{2} ln (x^{2} + a^{2})$ and $\frac{C}{x ^{2} + a ^{2}}$ as $\frac{C}{a} arctan \frac{x}{a}$ .

In practiceFind $\int \frac{5 x ^{2} + 4 x + 3}{( x + 1 ) ( x ^{2} + 1 )} d x$ .

$\frac{5 x ^{2} + 4 x + 3}{( x + 1 ) ( x ^{2} + 1 )} = \frac{A}{x + 1} + \frac{B x + C}{x ^{2} + 1}$ a linear numerator over the quadratic
$5 x^{2} + 4 x + 3 = A (x^{2} + 1) + (B x + C) (x + 1)$ multiply through
$x = - 1 : 4 = 2 A ⟹ A = 2$
$x^{2} : 5 = A + B ⟹ B = 3; x^{0} : 3 = A + C ⟹ C = 1$ check the $x$ terms: $B + C = 4$
$\int = 2 ln ∣ x + 1∣ + \frac{3}{2} ln (x^{2} + 1) + arctan x + c$

Notes

If the numerator's degree is not less than the denominator's, divide first.
A common error is to give the quadratic factor only a constant numerator. Its numerator must be one degree lower than it: linear.

Worked example

Express:

\frac{5 x ^{2} + 3 x + 4}{( x + 1 ) ( x ^{2} + 1 )}

in partial fractions, and hence show that:

\int_{0}^{1} \frac{5 x ^{2} + 3 x + 4}{( x + 1 ) ( x ^{2} + 1 )} d x = \frac{π}{4} + 4 ln 2

Show worked solution

Let:

5 x^{2} + 3 x + 4 = A (x^{2} + 1) + (B x + C) (x + 1)

At $x = - 1$ : $6 = 2 A$ , so $A = 3$ .

Comparing $x^{2}$ : $5 = A + B$ , so $B = 2$ .

Constants: $4 = A + C$ , so $C = 1$ .

So the integrand is:

\frac{3}{x + 1} + \frac{2 x + 1}{x ^{2} + 1}

and the integral is:

[3 ln (x + 1) + ln (x^{2} + 1) + arctan x]_{0}^{1} = 3 ln 2 + ln 2 + \frac{π}{4} = \frac{π}{4} + 4 ln 2

4.08

The mean value of a function

The mean value of a function 4.08

Key results

The mean value of $f (x)$ over $a \leq x \leq b$ is $\overset{ˉ}{f} = \frac{1}{b - a} \int_{a}^{b} f (x) d x$ .
$\overline{f + k} = \overset{ˉ}{f} + k$ and $\overline{k f} = k \overset{ˉ}{f}$ .

Notes

Geometrically, $\overset{ˉ}{f}$ is the height of the rectangle on $[a, b]$ with the same (signed) area as the region under the curve.
The mean of $sin x$ over a whole period is 0, because the areas above and below the axis cancel; over $[0, π]$ it is $\frac{2}{π}$ .
Physics uses this constantly: the mean power of an alternating current is the mean value of $I^{2} R$ over a cycle.

4.08

Maclaurin series

Maclaurin series 4.08

Key results

$f(x)=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^2+\dots+\dfrac{f^{(r)}(0)}{r!}x^r+\dots$ , valid for the values of $x$ where the series converges.
$e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \dots$ for all $x$ .
$sin x = x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \dots$ and $cos x = 1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \dots$ for all $x$ (in radians).
$ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ for $- 1 < x \leq 1$ .
$(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$ for $∣ x ∣ < 1$ (any real $n$ ).

Method

Differentiate repeatedly, evaluate each derivative at $x = 0$ , and substitute into the formula.

In practiceFind the Maclaurin series of $ln (cos x)$ up to the term in $x^{4}$ .

$\displaystyle f=\ln\cos x,\quad f'=-\tan x,\quad f''=-\sec^2x$
$\displaystyle f'''=-2\sec^2x\tan x,\quad f^{(4)}=-2\sec^4x-4\sec^2x\tan^2x$ product and chain rules
$\displaystyle f(0)=0,\ f'(0)=0,\ f''(0)=-1,\ f'''(0)=0,\ f^{(4)}(0)=-2$ evaluate at $x = 0$
$ln cos x = - \frac{x ^{2}}{2 !} - \frac{2 x ^{4}}{4 !} + \dots = - \frac{x ^{2}}{2} - \frac{x ^{4}}{12} - \dots$ valid for $∣ x ∣ < \frac{π}{2}$

Notes

A Maclaurin polynomial matches the function's value and first few derivatives at $x = 0$ ; it is accurate near 0 and can be poor far from it, as the sheet shows for $sin x$ .
$f$ must be defined and differentiable at 0, which is why $ln x$ has no Maclaurin series and $ln (1 + x)$ is used instead.

Building series from standard ones 4.08

Method

Substitute: replace $x$ by $2 x$ , $- x$ or $x^{2}$ in a standard series. The validity range changes accordingly: $ln (1 + 2 x)$ needs $- 1 < 2 x \leq 1$ .
Multiply two series, keeping only the terms up to the power required.
Differentiate or integrate a series term by term within its range of validity.

In practiceFind the series of $e^{x} ln (1 + x)$ up to the term in $x^{3}$ .

$e^{x} = 1 + x + \frac{x ^{2}}{2} + \dots, ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ standard series
$(1 + x + \frac{x ^{2}}{2}) (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3})$ keep only terms that can reach $x^{3}$
$= x + (- \frac{1}{2} + 1) x^{2} + (\frac{1}{3} - \frac{1}{2} + \frac{1}{2}) x^{3} + \dots$ collect each power
$e^{x} ln (1 + x) = x + \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + \dots$ valid for $- 1 < x \leq 1$

Notes

Multiplying series is usually much quicker than repeated differentiation: $e^{x} sin x = (1 + x + \frac{x ^{2}}{2} + \dots) (x - \frac{x ^{3}}{6} + \dots) = x + x^{2} + \frac{x ^{3}}{3} + \dots$
Series give limits directly: $\frac{sin x}{x} = 1 - \frac{x ^{2}}{6} + \dots \to 1$ as $x \to 0$ .

Worked example

Find the Maclaurin series of $ln (1 + x)$ up to the term in $x^{4}$ , and use it to estimate $ln 1.1$ .

Show worked solution

$f (0) = 0$ ;

f&#039;(x)=(1+x)^{-1}

f&#039;(0)=1

f&#039;&#039;(x)=-(1+x)^{-2}

f&#039;&#039;(0)=-1

f&#039;&#039;&#039;(0)=2

f^{(4)} (0) = - 6

So:

ln (1 + x) = x - \frac{x ^{2}}{2} + \frac{2 x ^{3}}{3 !} - \frac{6 x ^{4}}{4 !} + \dots = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \frac{x ^{4}}{4} + \dots

With $x = 0.1$ :

0.1 - 0.005 + 0.000333 - 0.000025 = 0.095308

against the true value $0.095310$ .

4.08

Improper integrals

Improper integrals 4.08

Definitions

Improper integral: an integral over an infinite interval, or of a function that is undefined (unbounded) at a point in the interval of integration.
Convergent: the defining limit exists and is finite. Divergent: it does not.

Key results

$\int_{a}^{\infty} f (x) d x = t \to \infty lim \int_{a}^{t} f (x) d x$ .
If $f$ is unbounded at $x = a$ : $\int_{a}^{b} f (x) d x = t \to a^{+} lim \int_{t}^{b} f (x) d x$ .
$\int_{1}^{\infty} \frac{d x}{x ^{p}}$ converges if and only if $p > 1$ ; $\int_{0}^{1} \frac{d x}{x ^{p}}$ converges if and only if $p < 1$ .
If both limits are infinite, or the integrand is unbounded inside the interval, split at a convenient point and require every part to converge: $\int_{- \infty}^{\infty} = \int_{- \infty}^{0} + \int_{0}^{\infty}$ .

Method

Replace the infinite limit (or the troublesome point) by $t$ , integrate as usual, then find the limit as $t$ tends to infinity (or to the point).

In practiceEvaluate $\int_{0}^{\infty} x e^{- 2 x} d x$ .

$\int_{0}^{\infty} x e^{- 2 x} d x = t \to \infty lim \int_{0}^{t} x e^{- 2 x} d x$ replace $\infty$ by $t$
$\int_{0}^{t} x e^{- 2 x} d x = [- \frac{1}{2} x e^{- 2 x} - \frac{1}{4} e^{- 2 x}]_{0}^{t}$ integration by parts
$= - \frac{1}{2} t e^{- 2 t} - \frac{1}{4} e^{- 2 t} + \frac{1}{4}$
$\to \frac{1}{4} as t \to \infty$ $t e^{- 2 t} \to 0$ and $e^{- 2 t} \to 0$

Notes

Show the limit explicitly in the working. "Substituting $\infty$ " is not acceptable, and loses marks.
Useful limits: $e^{- t} \to 0$ and $t e^{- t} \to 0$ as $t \to \infty$ ; $t ln t \to 0$ as $t \to 0^{+}$ (exponentials beat powers, powers beat logarithms).
The two curves on the sheet look alike, yet the area under $\frac{1}{x ^{2}}$ is finite and the area under $\frac{1}{x}$ is not: whether an integral converges depends on how fast the integrand decays.
$\int_{- 1}^{1} \frac{d x}{x ^{2}}$ diverges, because $\int_{0}^{1} \frac{d x}{x ^{2}}$ does. Substituting the limits blindly gives $[- \frac{1}{x}]_{- 1}^{1} = - 2$ : a negative area for a positive integrand, which shows that something has gone wrong.

Worked example

Show that:

\int_{0}^{\infty} x e^{- x} d x

converges and find its value.

Show worked solution

By parts,

\int_{0}^{t} x e^{- x} d x = [- x e^{- x}]_{0}^{t} + \int_{0}^{t} e^{- x} d x = - t e^{- t} + 1 - e^{- t}

As $t \to \infty$ , $t e^{- t} \to 0$ and $e^{- t} \to 0$ , so the limit exists and equals 1.

The integral converges to 1.

4.08

Volumes of revolution

Volumes of revolution 4.08

Key results

Rotating the region under $y = f (x)$ , $a \leq x \leq b$ , through $2 π$ about the $x$ -axis: $V = π \int_{a}^{b} y^{2} d x$ .
Rotating the region between a curve and the $y$ -axis, $c \leq y \leq d$ , about the $y$ -axis: $V = π \int_{c}^{d} x^{2} d y$ .
For a curve given parametrically: $V = π \int y^{2} \frac{d x}{d t} d t$ with limits in $t$ .

Method

Sketch the region and the axis of rotation.
For rotation about the $y$ -axis, rearrange to make $x^{2}$ a function of $y$ and use $y$ -limits.
For a region between two curves (a solid with a hole), subtract: $V = π \int (y_{outer}^{2} - y_{inner}^{2}) d x$ .

In practiceThe region between $y = x^{2}$ and $y = 2 x$ is rotated through $2 π$ about the $x$ -axis. Find the volume.

$x^{2} = 2 x ⟹ x = 0, 2$ limits from the intersections
$V = π \int_{0}^{2} ((2 x)^{2} - (x^{2})^{2}) d x$ outer curve $y = 2 x$ , inner $y = x^{2}$
$= π [\frac{4}{3} x^{3} - \frac{1}{5} x^{5}]_{0}^{2} = π (\frac{32}{3} - \frac{32}{5})$
$= \frac{64 π}{15}$

Notes

Each slice of width $δ x$ is approximately a disc of volume $π y^{2} δ x$ ; the integral is the limit of the sum of the discs.
Square $y$ before integrating: $π \int y^{2} d x$ is not $π (\int y d x)^{2}$ .
Give an exact answer as a multiple of $π$ unless told otherwise.

Worked examples

Worked example

The region enclosed by $y = 4 - x^{2}$ and the $x$ -axis is rotated through $2 π$ about the $x$ -axis.

Find the exact volume.

Show worked solution

The curve meets the axis at $x = \pm 2$ .

V = π \int_{- 2}^{2} (4 - x^{2})^{2} d x = π \int_{- 2}^{2} (16 - 8 x^{2} + x^{4}) d x = π [16 x - \frac{8 x ^{3}}{3} + \frac{x ^{5}}{5}]_{- 2}^{2} = 2 π (32 - \frac{64}{3} + \frac{32}{5}) = \frac{512 π}{15}

Worked example

The region between $y = x^{2}$ , the $y$ -axis and the line $y = 4$ (with $x \geq 0$ ) is rotated through $2 π$ about the $y$ -axis.

Find the volume.

Show worked solution

About the $y$ -axis,

V = π \int_{0}^{4} x^{2} d y

and $x^{2} = y$ .

So:

V = π \int_{0}^{4} y d y = π [\frac{y ^{2}}{2}]_{0}^{4} = 8 π