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C4 Cheat Sheet
Chapter Usual types of questions Tips What can go ugly 1 – Partial Fractions
Be able to split a fraction whose denominator is a product of linear expressions, e.g.
2 𝑥+ 3 𝑥(𝑥+ 1 ) Be able to split a fraction where one (or more) of the factors in the denominator are squared, e.g.
2 𝑥+ 3 𝑥^2 (𝑥+ 1 ) Deal with top-heavy fractions where the highest power in the denominator is greater or equal to the highest power in the denominator, e.g
𝑥^2 + 2 𝑥(𝑥+ 1 )
The textbook provides two methods for dealing with top heavy fractions. The algebraic long division method is miles easier! e.g.
𝑥^2 + 2 𝑥(𝑥+ 1 ) =^
𝑥^2 + 2 𝑥^2 +𝑥. Using long division we get^ a quotient of 1 and a remainder of −𝑥 + 2 , thus: 𝑥^2 + 2 𝑥^2 + 𝑥
Then split the
−𝑥+ 2 𝑥(𝑥+ 1 ) into partial fractions as normal. Don’t forget that when you have a squared factor in the denominator, you need two fractions in your partial fraction sum: 2 𝑥^2 (𝑥 + 1 )
𝑥^2
When you have three unknowns it’s generally easiest to use substitution to get two of them (e.g. the 𝐴 and the 𝐵) then compare the coefficients of 𝑥^2 to get the 𝐶. For the above example: 2 ≡ 𝐴𝑥(𝑥 + 1 )^ + 𝐵(𝑥 + 1 )^ + 𝐶𝑥^2 We can see immediately, without needing to write out the expansion, that 0 = 𝐴 + 𝐶, by comparing 𝑥^2 terms.
Forgetting the extra term when the denominator’s factors are squared. Being sloppy at algebraic long division! Be careful with substitution of negative values. You may have to spot that you need to factorise the denominator first before expressing as partial fractions. Not realising the fraction is top heavy and therefore trying to incorrectly do: 2 𝑥^2 𝑥(𝑥 + 1 )^
2 – Parametric Equations ^ Know that^
𝑑𝑦 𝑑𝑥 =^
(𝑑𝑦 𝑑𝑡) (𝑑𝑥 𝑑𝑡) (This makes sense as we have just divided numerator and denominator by 𝑑𝑡) Be able to integrate parametric equations. Be able to convert parametric equations into a single Cartesian one.
Note: You will NOT be asked to sketch parametric equations. To convert parametric equations involving trig functions to Cartesian ones, the strategy is usually to make sin 𝑥 and cos 𝑥 the subject before using the identity sin^2 𝑥 + cos^2 𝑥 ≡
- Often squaring one of the parametric equations helps so that we have sin^2 𝑥 and/or cos^2 𝑥: 𝑥 = √ 3 sin 2 𝑡 𝑦 = 4 cos^2 𝑡 𝑥 = 2 √ 3 sin 𝑡 cos 𝑡 𝑥^2 = 12 sin^2 𝑡 cos^2 𝑡 𝑥^2 = 12 ( 1 − cos^2 𝑡) cos^2 𝑡 𝑥^2 = 12 ( 1 − (
𝑦 4
)
2 )
𝑦 4
Hitting a dead end converting parametric equations to Cartesian. See tips on left. Forgetting to multiply by 𝑑𝑥 𝑑𝑡 when integrating parametric equations. Remember that the 𝑑𝑥 in (^) ∫ 𝑦 𝑑𝑥 can be replaced with 𝑑𝑥 𝑑𝑡 𝑑𝑡, which is easy to remember, as the 𝑑𝑡’s cancel if we think of 𝑑𝑥 and 𝑑𝑡 just as quantities.
3 – Binomial Expansion
Expanding out an expression of the form ( 1 + 𝑘𝑥)𝑛, where 𝑛 is negative or fractional. Expanding out an expression of the form (𝑎 + 𝑘𝑥)𝑛, where 𝑎 needs to be factorised out first. Finding the product of two Binomial expansions, e.g. √ 1 + 𝑥 √^1 −^ 𝑥^
1 (^2) ( 1 − 𝑥)−
1 2
( 1 + 𝑘𝑥)𝑛^ = 1 + 𝑛(𝑘𝑥)^ +
𝑛(𝑛− 1 ) 2! (𝑘𝑥)
2
(𝑘𝑥)^3 + ⋯
Your expression may be a binomial expansion in disguise, e.g. 1 √^1 −^2 𝑥^
= ( 1 − 2 𝑥)−
1 (^2) = 1 + (− 1 2 )^
(− 2 𝑥) (^) +
− 12 × − (^32) 2!
(− 2 𝑥)^2 + ⋯
When the first term is not 1, you have to factorise this number out, raised to the power outside the brackets. e.g.
1 (^2) = 4
1 (^2) ( 1 +
1 2
= 2 [ 1 +
𝑥) + ⋯ ]
Ensure the outer brackets are maintained till the very end, when you expand them out. When finding the product of two expansions, then if you needed up to the 𝑥^2 term, then you only need to find up to the 𝑥^2 term in each of the two expansions. Only consider things in the expansion which are up to 𝑥^2. e.g.
√^1 +^ 𝑥 1 − 𝑥 =^
√^1 +^ 𝑥 √ 1 − 𝑥
= ( 1 + 𝑥)
1 (^2) ( 1 − 𝑥)−
1 2
( 1 + 𝑥)
1 (^2) ≈ 1 + 1 2 𝑥^ −^
1 8 𝑥
2
( 1 − 𝑥)−
1 (^2) ≈ 1 + 1 2 𝑥^ +^
3 8 𝑥
2
( 1 + 𝑥)
1 (^2) ( 1 − 𝑥)−
1 (^2) ≈ ( 1 + 1 2 𝑥^ −^
1 8 𝑥
(^2) ) ( 1 + 1 2 𝑥^ +^
3 8 𝑥
(^2) )
= 1 +
1 2 𝑥^ +^
3 8 𝑥
(^2) + 1 2 𝑥^ +^
1 4 𝑥
(^2) − 1 8 𝑥
2
= 1 + 𝑥 +
1 2 𝑥
2
Many things! Lack of brackets when squaring/cubing things, e.g. you need ( 2 𝑥)^3 = 8 𝑥^3 not 2 𝑥^3 With say ( 3 + 4 𝑥)−^1 , forgetting to raise the 3 you factor out to the power of -1. Forgetting to put the factorial in the denominators of the Binomial coefficients (a common error is
□ 3 instead of^
□ 3 !) Being careless in using your calculator when simplifying coefficients. Be ridiculously careful with signs! Accidentally forgetting the minus in the power when expanding say (^) (𝑥+^11 ) 2
5 - Vectors (In rough descending order of how frequently they appear in exams) Find the point of intersection of two lines or prove that two lines do not intersect. Find the angle between two lines. Finding a missing 𝑥/𝑦/𝑧 value of a point on a line. Find the length of a vector or the distance between two points. Find the nearest point on a line to a point not on the line (often the origin) – note: not in your textbook! Show lines are perpendicular. Show a point lies on a line. Show 3 points are collinear (i.e. lie on the same straight line) Find the area of a rectangle, parallelogram or triangle formed by vectors. Find the equation of a line. Find the reflection of a point in a line.
When you see the 𝑖, 𝑗, 𝑘 unit vectors used in an exam question, never actually use this notation yourself: always just write all vectors in conventional column vector form. Almost always draw a suitable diagram. This will be particularly helpful when you need to find the area of some shape (typically the last part of a question). When finding the area of a shape, you can almost always use your answers from previous parts of the questions, including lengths of vectors and angles between two vectors. Remember that area of non-right angled triangle = 12 𝑎𝑏 sin 𝐶 where the angle 𝐶 appears between the two sides 𝑎 and 𝑏. A parallelogram can be cut in half to form two congruent non- right angled triangles (i.e. multiply by 2). To show 3 points 𝐴, 𝐵, 𝐶 are collinear, just show that 𝐴𝐵⃗⃗⃗⃗⃗ is a multiple of 𝐵𝐶⃗⃗⃗⃗⃗ (i.e. vectors are parallel). “Show 3 𝒊 + 3 𝒋 + 2 𝒌 lies on the line with vector equation 𝑟 = 𝒊 + 3 𝒋 + 4 𝒌 + 𝑡(𝒊 − 𝒌)”
i.e. Show (
) lies on (
). Equating 3 = 1 + 𝑡 to 𝑡 = 2.
Then 4 − 𝑡 = 4 − 2 = 2 , so 𝑦 and 𝑧 components are same. “Let 𝑙 1 : 𝒓 = ( 9 𝒊 + 13 𝒋 − 3 𝒌)^ + 𝜆(𝒊 + 4 𝒋 − 2 𝒌) Given point 𝐴 has positive vector 4 𝑖 + 16 𝑗 − 3 𝑘 and 𝑃 lies on 𝑙 1 such that 𝐴𝑃 is perpendicular to 𝑙 1 , find 𝑃.”
𝑙 1 : (
Note that the direction vector of the line, and the vector 𝑃𝐴⃗⃗⃗⃗⃗ are perpendicular. 𝑃 is just a point on the line so can be
represented as (
) for some specific 𝜆 we need to
find.
Direction vector of 𝑙 1 is (
When finding the angle between two lines, accidentally using the full vector representation of the line (in your dot product), and not just the direction
component, e.g. using (
) instead of
just the correct (
Making sign errors when subtracting vectors, particularly when subtracting an expression involving a negative. Correctly:
(
Once finding out 𝑠 and 𝑡 (or 𝜇 and 𝜆) when solving simultaneous equation to find the intersection of two lines, forgetting to show that these satisfy the remaining equation. Forgetting the square root when finding the magnitude of a vector.
𝑃𝐴⃗⃗⃗⃗⃗ = (
4 16 − 3
) − (
9 + 𝜆 13 + 4 𝜆 − 3 − 2 𝜆
) = (
− 5 − 𝜆 3 − 4 𝜆 2 𝜆
)
Thus: (
− 5 − 𝜆 3 − 4 𝜆 2 𝜆
) ⋅ (
1 4 − 2
) = 0
− 5 − 𝜆 + 12 − 16 𝜆 − 4 𝜆 = 0 𝜆 =
1 3
Thus: 𝑃 =
(
9 + (^13) 13 + 4 (^13 ) − 3 − 2 (^13 ))
=
(
(^9 ) (^14 ) − (^3 23) )
6 - Integration (^) Integrate a large variety of expressions. See the ‘integration cheat sheet’ overleaf. But by category: o Integrating trig functions, including reciprocal functions and squared functions sin^2 𝑥, cos^2 𝑥 , sec^2 2 𝑥, etc. o Integrating by ‘reverse chain rule’ (also known as ‘integration by inspection’). o Integrating by a given substitution. o Integration by parts. o Integrating by use of partial fractions. o Integrating top heavy fractions by algebraic division.
One often forgotten integration is exponential functions such as 2 𝑥. Differentiating has effect of multiplying by 𝑙𝑛 of the base, and thus integrating divides by it. i.e. 𝑑 𝑑𝑥
( 2 𝑥) (^) = ln 2 ⋅ 2 𝑥^ ∫ 2 𝑥^ 𝑑𝑥 =
1 ln 2 2
𝑥 (^) + 𝐶
Know the two double angle formulae for 𝑐𝑜𝑠 like the back of your hand, for use when integrating sin^2 𝑥 or cos^2 𝑥 In general, know your integrals of all the ‘trig squares’, i.e. sin^2 𝑥 , cos^2 𝑥 , tan^2 𝑥 , 𝑐𝑜𝑠𝑒𝑐^2 𝑥, sec^2 𝑥 , cot^2 𝑥 For integration by ‘reverse chain rule’, always ‘consider’ some sensible expression to differentiate, then adjust for the factor difference. e.g. ∫( 4 − 3 𝑥)^5 𝑑𝑥 Then your working might be: “Consider 𝑦 = ( 4 − 3 𝑥)^6. Then 𝑑𝑦 𝑑𝑥
= 6 ( 4 − 3 𝑥)^5 × (− 3 ) = − 18 ( 4 − 3 𝑥)^5
∴ ∫( 4 − 3 𝑥)^5 𝑑𝑥 = −
1 18
( 4 − 3 𝑥)^6 + 𝐶 For integration by substitution, the official specification says “Except in the simplest of cases, the substitution will be given.” Remember that starting with the substitution, say 𝑢 = 𝑥^2 + 1 , it helps to make 𝑥 the subject, except in some cases where there’s a trigonometric substitution, e.g. if 𝑢 = sin 𝑥 + 1 , but sin 𝑥 appears in the expression to integrate, then we might
Where to start! One big problem is just not knowing what method to use to integrate a particular expression. The cheat sheet overleaf should help, as should lots of practice of a variety of expressions! Similarly getting stuck on integration by substitution, because you can’t get the whole original expression only in terms of the new variable (𝑡 or otherwise). Perhaps the all-time biggest mistake is forgetting to consider the effects of chain rule. e.g. Accidentally doing
∫ cos 2 𝑥 𝑑𝑥 = sin 2 𝑥 Sign errors when integrating/differentiating trig functions. Other than sin and cos, be careful about cot/cosec: 𝑑 𝑑𝑥
(cot 𝑥) (^) = −𝑐𝑜𝑠𝑒𝑐^2 𝑥
thus ∫ 𝑐𝑜𝑠𝑒𝑐^2 𝑥 𝑑𝑥 = − cot 𝑥 + 𝐶 A common one: Forgetting about the chain rule when integrating expressions of the form (𝑎 + 𝑏𝑥)𝑐, see ∫( 4 − 3 𝑥)^5 𝑑𝑥 example.
Note the nice double negative tidying up trick towards the end.
If you’re solving
𝑑𝑦 𝑑𝑥 =^ 𝑥𝑦^ +^ 𝑦, then you need the^ 𝑦^ (or whatever variable appears at the top of
𝑑𝑦 𝑑𝑥) on the LHS. This is always achieved by a division or multiplication, which may require factorisation first: 𝑑𝑦 𝑑𝑥
𝑑𝑦 = ∫ 𝑥 + 1 𝑑𝑥 ln|𝑦| =
𝑥^2 + 𝑥 + 𝐶
1 2 𝑥
(^2) +𝑥+𝐶 = 𝐴𝑒
1 2 𝑥
(^2) +𝑥
Note in the above example, we let some new constant 𝐴 = 𝑒𝐶^ to help tidy things up. If we had ln 𝑥 + 𝐶 on the right- hand-side, we’d make 𝐶 = ln 𝐴 so that ln 𝑥 + ln 𝐴 = ln(𝐴𝑥). Similarly if we had ln 𝑦 = 𝑥 + 𝐶, and hence 𝑦 = 𝑒𝑥+𝐶^ = 𝑒𝑥𝑒𝐶, we could make 𝐴 = 𝑒𝐶. In differential equations, ensure you separate the RHS into the form 𝑓(𝑥)𝑔(𝑦) first so that you are able to divide by 𝑔(𝑦), e.g. 𝑑𝑦 𝑑𝑥 = 𝑥 + 𝑥𝑦 → 𝑑𝑦 𝑑𝑥 = 𝑥( 1 + 𝑦) → (^1) +^1 𝑦 𝑑𝑦 = 𝑥 𝑑𝑥
In differential equations, if you’re given initial conditions (note, often 𝑡 = 0 is often implied for the initial condition), then it’s generally easier to plug them in to work out your constant of integration sooner rather than later.
C4 Integration Cheat Sheet
𝒇(𝒙) How to deal with it (^) ∫ 𝒇(𝒙)𝒅𝒙 (+constant) FormBk? 𝒔𝒊𝒏 𝒙 Standard result − cos 𝑥 No 𝒄𝒐𝒔 𝒙 Standard result sin 𝑥 No 𝒕𝒂𝒏 𝒙 In formula booklet, but use ∫ sin cos^ 𝑥 𝑥 𝑑𝑥 which is of the form ∫ 𝑘𝑓
′(𝑥) 𝑓(𝑥) 𝑑𝑥
ln|sec 𝑥| Yes
𝒔𝒊𝒏𝟐^ 𝒙 For both sin^2 𝑥 and cos^2 𝑥 use identities for cos 2 𝑥 cos 2 𝑥 = 1 − 2 sin^2 𝑥 sin^2 𝑥 =
1 2
−
1 2
cos 2 𝑥
1 2
𝑥 −
1 4
sin 2 𝑥 No
𝒄𝒐𝒔𝟐^ 𝒙 cos 2 𝑥 = 2 cos^2 𝑥 − 1 cos^2 𝑥 =
1 2
1 2
cos 2 𝑥
1 2
𝑥 +
1 4
sin 2 𝑥 No
𝒕𝒂𝒏𝟐𝒙 1 + tan^2 𝑥 ≡ sec^2 𝑥 tan^2 𝑥 ≡ sec^2 𝑥 − 1
tan 𝑥 − 𝑥 No
𝒄𝒐𝒔𝒆𝒄 𝒙 Would use substitution 𝑢 = 𝑐𝑜𝑠𝑒𝑐 𝑥 + cot 𝑥, but too hard for exam.
−ln|𝑐𝑜𝑠𝑒𝑐 𝑥 + cot 𝑥| Yes
𝒔𝒆𝒄 𝒙 Would use substitution 𝑢 = sec 𝑥 + tan 𝑥, but too hard for exam.
ln|sec 𝑥 + tan 𝑥| Yes
𝒄𝒐𝒕 𝒙 (^) ∫ cos sin 𝑥𝑥 𝑑𝑥 which is of the form
∫ 𝑓
′(𝑥) 𝑓(𝑥) 𝑑𝑥
ln|𝑠𝑖𝑛 𝑥| Yes
𝒄𝒐𝒔𝒆𝒄𝟐𝒙 By observation. −^ cot^ 𝑥^ No! 𝒔𝒆𝒄𝟐𝒙 By observation.^ tan^ 𝑥^ Yes^ (but memorise) 𝒄𝒐𝒕𝟐𝒙 1 + cot^2 𝑥 ≡ 𝑐𝑜𝑠𝑒𝑐^2 𝑥 −^ cot^ 𝑥^ −^ 𝑥^ No 𝒆𝒙^ Standard result 𝑒𝑥^ No 𝒂𝒙^ 𝑦 = 𝑎𝑥^ → ln 𝑦 = 𝑥 ln 𝑎 Then differentiate implicitly.
1 ln(𝑎)
𝑎𝑥^ No 𝟏 𝒙
Standard result ln 𝑥 No
𝐥𝐧 𝒙 (^) Use IBP, where 𝑢 = ln 𝑥 , 𝑑𝑣 𝑑𝑥 = 1 𝑥 ln 𝑥 − 𝑥 No
𝒇(𝒙) How to deal with it (^) ∫ 𝒇(𝒙)𝒅𝒙 (+constant) FB? 𝒔𝒊𝒏 𝟐𝒙 𝒄𝒐𝒔 𝟐𝒙 For any product of sin and cos with same coefficient of 𝑥, use double angle. sin 2 𝑥 cos 2 𝑥 ≡ 12 sin 4 𝑥
−
1 8 cos^4 𝑥^
No
𝐜𝐨𝐬 𝒙 𝒆𝒔𝒊𝒏^ 𝒙^ Of form 𝑔′(𝑥)𝑓′(𝑔(𝑥)) 𝑒sin^ 𝑥 𝒙 𝒙 + 𝟏
Use algebraic division. 𝑥 𝑥 + 1 ≡^1 −^
1 𝑥 + 1
𝑥 − ln|𝑥 + 1 |
𝟏 𝒙(𝒙 + 𝟏)
Use partial fractions. ln|𝑥|^ − ln|𝑥 + 1 |
𝟒𝒙 𝒙𝟐^ + 𝟏
Reverse chain rule. Of form ∫ 𝑘𝑓
′(𝑥) 𝑓(𝑥)
2 ln|𝑥^2 + 1 |
𝒙 (𝒙𝟐^ + 𝟏)𝟐
Power around denominator so NOT of form (^) ∫ 𝑘𝑓
′(𝑥) 𝑓(𝑥). Rewrite as product. 𝑥(𝑥^2 + 1 )−^2 Reverse chain rule (i.e. “Consider 𝑦 = (𝑥^2 + 1 )−^1 " and differentiate.
−
1 2
(𝑥^2 + 1 )−^1
𝟖𝒙𝟐 𝟒𝒙𝟐^ − 𝟏
Fraction top heavy so do algebraic division first. Then split into algebraic fractions as 4 𝑥^2 − 1 = ( 2 𝑥 + 1 )( 2 𝑥 − 1 )
2 𝑥 + 12 ln| 1 − 2 𝑥|
−
1 2 ln^ |^2 𝑥^ +^1 | 𝒆𝟐𝒙+𝟏 𝟏 𝟏 − 𝟑𝒙
For any function where ‘inner function’ is linear expression, divide by coefficient of 𝑥
1 2 𝑒
2 𝑥+ 1
−
1 3 ln|^1 −^3 𝑥| 𝒙√𝟐𝒙 + 𝟏 Use sensible substitution.^ 𝑢^ = 2 𝑥 + 1 or even better, 𝑢^2 = 2 𝑥 + 1.
1 15
( 2 𝑥 + 1 )
3 2 ( 3 𝑥 − 1 )
𝐬𝐢𝐧𝟓^ 𝒙 𝒄𝒐𝒔 𝒙 Reverse chain rule.^1 6 sin
(^6) 𝑥 𝒔𝒊𝒏 𝟑𝒙 𝒄𝒐𝒔 𝟐𝒙 Use identities in C3 formula booklet, sin 3 𝑥 cos 2 𝑥 = 12 (sin 5 𝑥 + cos 3 𝑥) Note: has never come up in an exam.
−
1 10
cos 5 𝑥
1 6 sin^3 𝑥
Sort of
