Ch.11%20Integration

Oct 13, 2026

Integration is presented as the inverse of differentiation and as a toolkit for finding areas between curves and coordinate axes, with applications extending to Physics, Statistics, and Economics. The guide consolidates key standard results, including power, exponential, logarithmic, and trigonometric integrals, and notes derivation from the differentiation section. It then explains how to handle non-straightforward integrals by algebraic manipulation and identities, including example-driven approaches, partial fractions, and the trapezium rule for numerical area estimates. Integration via substitution is introduced as a method for more complex integrals.

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Very useful results

Standard integration results and derivations

Using integration for areas

Parametric curves and limits

Manipulating integrands

Trigonometric identities and examples

Techniques for harder integrals

Standard forms and constant adjustment

Partial fractions

Trapezium rule and error

Integration via substitution

Integration Cheat Sheet Integration is the inverse of differentiation. We can think of integration as a mathematical tool that allows us to find areas enclosed between curves and the coordinate axes. In fact, the uses of integration extend far beyond finding areas and can also be found in other fields of study, including Physics, Statistics and even Economics. Very useful results f(􀝔) 􀝔n eax+b 1 a􀯫+b (a􀝔+b)n sin k􀝔 cos k􀝔 ∫f(􀝔) d􀝔 􀯫􀳙+1 􀯡+1 1a eax+b 1a ln|a􀝔+b| 1 ሺax+bሻ􀝊+1 a n+1 − 1k cos k􀝔 1k sin k􀝔 These can be derived from the differentiation section in the formula booklet f(􀝔) tan k􀝔 cot k􀝔 cosec k􀝔 sec k􀝔 sec2 k􀝔 sec k􀝔 tan k􀝔 cosec2 k􀝔 cosec k􀝔 cot k􀝔 ∫f(􀝔) d􀝔 1􀯞 ln|sec k􀝔| 1􀯞 ln|sin k􀝔|− 1􀯞 ln|cosec k􀝔 + cot k􀝔| 1􀯞 ln|sec k􀝔 + tan k􀝔| 1􀯞 tan k􀝔 1􀯞 sec k􀝔− 1􀯞 cot k􀝔 1 − ~~ ~~ 􀯞 cosec k􀝔 “by parts” formula: Trapezium rule: You will also be given these in the formula booklet. Just like in Pure Year 1, you will need to use integration to find areas. You should remember: ▪ ∫􀯔􀯕 􀝕 􀝀􀝔 represents the area bounded between the curve 􀝕 = 􀝂ሺ􀝔ሻ, the 􀝔-axis and the lines 􀝔 = 􀜽 and 􀝔 = 􀜾 . ▪ When a function is given parametrically, the area under the curve is given by ∫􀯔􀯕 ቂ􀝕 􀯗􀯫􀯗􀯧ቃ 􀝀􀝐 . Remember that 􀜽 and 􀜾 are limits given in terms of 􀝐, and the integration is done with respect to 􀝐 . ▪ You need to be able to recognise that lim 􀰋􀯫→0 􀜾∑ 􀝂ሺ􀝔ሻ􀟜􀝔􀝔 =􀜽 = 􀯕􀝂ሺ􀝔ሻ 􀝀􀝔 You might come across an expression involving trigonometric functions that you can’t integrate straight away using one of the results above. For example, ∫sin3 􀝔 d􀝔 can’t be found directly. To tackle problems of this sort, you need to manipulate the expression into a form you can integrate. This is why it is crucial you are familiar with all of the identities you encountered in chapters 6 and 7. Below is an example showcasing we do this in practice. Example 1: Evaluate ∫ sin2 􀝔 d􀝔 ∫ sin2 􀝔 􀝀􀝔 = ∫ ሺ1 − 􀜿􀝋􀝏 2􀝔ሻ􀝀􀝔 = ∫ 1 􀝀􀝔 − ∫ 􀜿􀝋􀝏 2􀝔 􀝀􀝔 .􀜤􀝑􀝐 􀜿􀝋􀝏2􀝔 = 2􀜿􀝋􀝏 2􀝔 − 1, 􀝏􀝋 􀜿􀝋􀝏 2􀝔 = 􀜿􀝋􀝏2􀝔2~~ ~~+~~ ~~1 ∴ ∫ 􀜿􀝋􀝏 2􀝔 􀝀􀝔 = 12 ∫ 􀜿􀝋􀝏2􀝔 + 1 􀝀􀝔 = 12 [12 􀝏􀝅􀝊2􀝔 + 􀝔] = 14 􀝏􀝅􀝊2􀝔 + 12 􀝔􀝏􀝋 ∫ 􀝏􀝅􀝊2 􀝔 􀝀􀝔 = 􀝔 − 14 􀝏􀝅􀝊2􀝔 − 12 􀝔 + 􀜿 using 􀝏􀝅􀝊2x = 1 − 􀜿􀝋􀝏 2x using the double angle formulae 􀜿􀝋􀝏2􀝔 = 2􀜿􀝋􀝏 2􀝔 − 1 Some complicated expressions can be integrated very easily if they are of one of the forms below: [1] ∫ 􀝂′ ሺ􀝔ሻ [􀝂ሺ􀝔ሻ]􀯡 􀝀􀝔 = 􀯙ሺ􀯫􀯡1+1 + 􀜿 . e.g. ∫ 􀝏􀝅􀝊400􀝔 􀜿􀝋􀝏􀝔 􀝀􀝔 = 􀯦􀯜􀯡440011􀯫 + 􀜿∫ 􀝔2√􀝔 3 + 4 􀝀􀝔 = 13 ∙ ~~ ~~ሺ􀯫34ሻ32~~ ~~ [2] ∫ 􀯙􀯙′ሺ􀯫ሺ􀯫ሻሻ 􀝀􀝔 = ln|􀝂ሺ􀝔ሻ| + 􀜿 e.g. ∫ 􀯫2242 􀝀􀝔 = ln|􀝔2 + 42| + 􀜿 ∫ 􀯦2􀯜􀯖􀯡􀯫􀯢􀯦􀯖2􀯢􀯫1 􀝀􀝔 = ∫ 21~~2~~􀯖􀯦􀯢􀯜􀯦􀯡22􀯫1 􀝀􀝔 = 12 ቂ − 14 􀝈􀝊|2􀜿􀝋􀝏2􀝔 + 1|ቃ + 􀜿 Remember that you must adjust for any variation in constants. In example 2 of rule 1, we had to multiply our answer by 13 since the differential of ሺ􀝔3 + 4ሻ5 is 3􀝔2 , not 􀝔2 . While it is true you do not necessarily need to know the above rules, it is still very worthwhile for you to take the time to learn to apply them because they can greatly simplify otherwise difficult integrals. You could also be asked to integrate an expression involving a fraction with more than one linear factor in the 11􀯫2 +14􀯫+5 denominator. Take for example, the expression ሺ􀯫+1ሻ2ሺ2􀯫+1ሻ . This cannot be integrated directly so we need to simplify it, which is where partial fractions come in handy. Using partial fractions, we can rewrite this as 4 − 2 + 3 􀯫+1 ሺ􀯫+1ሻ2 ሺ2􀯫+1ሻ . Now, each of the above terms can be integrated directly using one of the standard results on the left. Carrying out the integration: ∫ ሺ1􀯫21ሻ+21ሺ42􀯫􀯫51ሻ 􀝀􀝔 = ∫ 􀯫~~ ~~1 − ሺ􀯫1ሻ2 + ሺ2􀯫3+1ሻ 􀝀􀝔= 4 ln|􀝔 + 1| + 􀯫1 + 32 ln|2􀝔 + 1| +c Such questions are often worth upwards of 6 marks and are heavily reliant on your ability to use partial fractions well. The integration is mostly straight forward and does not require much extra work outside of applying the standard results, but it is the partial fractions procedure that can be a little tedious. In some cases, you might not be able to

FAQs

How is integration related to differentiation, and what is it used for?

Integration is the inverse of differentiation. It is used to find areas enclosed between curves and the coordinate axes, and the same ideas extend to other fields such as Physics, Statistics, and Economics.

What should you do if an integral cannot be found directly using the standard results?

Manipulate the integrand into a form you can integrate, which is why recognizing the identities from earlier chapters is crucial. The document demonstrates this with an example that rewrites the integrand using identities before integrating.

When can you use partial fractions and the trapezium rule?

Use partial fractions when you face a rational expression with multiple linear factors in the denominator, so the integrand can be decomposed into simpler terms. Use the trapezium rule when a function cannot be integrated algebraically to estimate the area under the curve numerically.

Oct 13, 2026

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