Exercises Given the gradient 𝑑𝑦 𝑑𝑥 , find the original function 𝑦. Q 𝒅𝒚 𝒅𝒙 = 𝒚= 1 𝑥 5 1 6 𝑥 6 +𝑐 2 10 𝑥 4 2 𝑥 5 +𝑐 3 − 𝑥 −2 𝑥 −1 +𝑐 4 𝑥 2 3 3 5 𝑥 5 3 +𝑐 5 6 𝑥 3 3 2 𝑥 4 +𝑐 6 −5 −5𝑥+𝑐 7 6𝑥 3 𝑥 2 +𝑐 8 −14 𝑥 −8 2 𝑥 −7 +𝑐 9 2 𝑥 −0.4 10 3 𝑥 0.6 +𝑐 10 5 𝑥 − 3 2 −10 𝑥 − 1 2 +𝑐 ? ? ? ? ? ? ? ? ? ?

Further Examples 2 𝑥 3 −3 𝑥 𝑑𝑥 = 2 𝑥 −3 −3 𝑥 1 2 = − 𝑥 −2 −2 𝑥 3 2 +𝑐 2𝑥 2 + 𝑥 +5 𝑥 2 𝑑𝑥 = 4 𝑥 2 + 𝑥 − 3 2 +5 𝑥 −2 𝑑𝑥 = 4 3 𝑥 3 − 2 𝑥 − 5 𝑥 +𝑐 Notice that at this point, we haven’t actually integrated yet, only simplified, hence the integral symbol remains! ? 𝑑𝑥 ?

Working out the c Suppose that for a curve 𝐶, 𝑑𝑦 𝑑𝑥 =2𝑥 and the curve passes through the point 3,10 . Work out 𝑦. By integrating, 𝒚= 𝒙 𝟐 +𝒄 Substituting: 𝟏𝟎= 𝟑 𝟐 +𝒄. So 𝒄=𝟏. 𝒚= 𝒙 𝟐 +𝟏 A curve 𝑦=𝑓 𝑥 passes through the point 4,5 and 𝑓 ′ 𝑥 = 𝑥 2 −2 𝑥 𝒇 ′ 𝒙 = 𝒙 𝟑 𝟐 −𝟐 𝒙 − 𝟏 𝟐 𝒇 𝒙 = 𝟐 𝟓 𝒙 𝟓 𝟐 −𝟒 𝒙 𝟏 𝟐 +𝒄 𝟓= 𝟐 𝟓 𝟒 𝟓 𝟐 −𝟒 𝟒 +𝒄 so 𝒄= 𝟏 𝟓 Thus 𝒚= 𝟐 𝟓 𝒙 𝟓 𝟐 −𝟒 𝒙 𝟏 𝟐 + 𝟏 𝟓 ? ?

Schoolboy ErrorsTM ? ? 𝑑𝑦 𝑑𝑥 = 2 𝑥 2 +3 =2 𝑥 −2 +3 =−2 𝑥 −1 +3𝑥+𝑐 What’s wrong with these workings? 𝑑𝑦 𝑑𝑥 = 2 𝑥 2 +3 =2 𝑥 −2 +3 =−2 𝑥 −1 +3𝑥+𝑐 𝑥 2 𝑑𝑥 = 1 3 𝑥 3 +𝑐 ? They forgot to put 𝑦= on the third line. ? They’ve integrated on the second line, so they don’t want the integral symbol!

Evaluating Definite Integrals 1 2 2 𝑥 3 +2𝑥 𝑑𝑥 = 1 2 𝑥 4 + 𝑥 2 1 2 = 8+4 − 1 2 +1 = 21 2 −2 −1 4𝑥 3 +3 𝑥 2 𝑑𝑥 = 𝑥 4 + 𝑥 3 −2 −1 = 1−1 − 16−8 =−8 ? ?

Harder Examples ? ? The Sketch The number crunching Sketch the curve with equation 𝑦=𝑥 𝑥−1 𝑥+3 and find the area between the curve and the 𝑥-axis. The Sketch The number crunching ? 𝑦 ? 𝑥 𝑥−1 𝑥+3 = 𝑥 3 −2 𝑥 2 −3𝑥 −3 0 𝑥 3 −2 𝑥 2 −3𝑥 𝑑𝑥 =11.25 0 1 𝑥 3 −2 𝑥 2 −3𝑥 𝑑𝑥 =− 7 12 Adding: 11.25+ 7 12 =11 5 6 𝑥 -3 1

Integration

Integration 𝑦= 𝑥 𝑛 𝑦=𝑠𝑖𝑛𝑥 𝑦=𝑐𝑜𝑠𝑒𝑐𝑥 𝑑𝑦 𝑑𝑥 =𝑛 𝑥 𝑛−1 𝑑𝑦 𝑑𝑥 =𝑐𝑜𝑠𝑥 You need to be able to integrate standard functions You met the following in C3, in the differentiation chapter: 𝑦= 𝑥 𝑛 𝑦=𝑠𝑖𝑛𝑥 𝑦=𝑐𝑜𝑠𝑒𝑐𝑥 𝑑𝑦 𝑑𝑥 =𝑛 𝑥 𝑛−1 𝑑𝑦 𝑑𝑥 =𝑐𝑜𝑠𝑥 𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥 𝑦= 𝑒 𝑓(𝑥) 𝑦=𝑐𝑜𝑠𝑥 𝑦=𝑐𝑜𝑡𝑥 𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥)𝑒 𝑓(𝑥) 𝑑𝑦 𝑑𝑥 =−𝑠𝑖𝑛𝑥 𝑑𝑦 𝑑𝑥 =−𝑐𝑜𝑠𝑒 𝑐 2 𝑥 𝑦=ln⁡(𝑓 𝑥 ) 𝑦=𝑡𝑎𝑛𝑥 𝑦=𝑠𝑒𝑐𝑥 𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥) 𝑓(𝑥) 𝑑𝑦 𝑑𝑥 =𝑠𝑒 𝑐 2 𝑥 𝑑𝑦 𝑑𝑥 =𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 6A

Integration 𝑥 𝑛 = 𝑥 𝑛+1 𝑛+1 +𝐶 𝑐𝑜𝑠𝑥= 𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥= 𝑠𝑖𝑛𝑥+𝐶 −𝑐𝑜𝑠𝑒𝑐𝑥+𝐶 You need to be able to integrate standard functions Therefore, you already can deduce the following 𝑥 𝑛 = 𝑥 𝑛+1 𝑛+1 +𝐶 𝑐𝑜𝑠𝑥= 𝑐𝑜𝑠𝑒𝑐𝑥𝑐𝑜𝑡𝑥= 𝑠𝑖𝑛𝑥+𝐶 −𝑐𝑜𝑠𝑒𝑐𝑥+𝐶 𝑒 𝑥 = 𝑠𝑖𝑛𝑥= 𝑐𝑜𝑠𝑒 𝑐 2 𝑥= 𝑒 𝑥 +𝐶 −𝑐𝑜𝑠𝑥+𝐶 −𝑐𝑜𝑡𝑥+𝐶 1 𝑥 = 𝑠𝑒 𝑐 2 𝑥= 𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥= ln⁡|𝑥|+𝐶 𝑡𝑎𝑛𝑥+𝐶 𝑠𝑒𝑐𝑥+𝐶 The modulus sign is used here to avoid potential problems with negative numbers… (More info on the next slide!) 6A

Integration 𝒚= 𝟏 𝒙 1 𝑥 = ln⁡|𝑥|+𝐶 𝒚= 𝟏 −𝒙 𝑓 𝑥 = 1 𝑥 𝑓 𝑥 = 1 −𝑥 6A You need to be able to integrate standard functions Therefore, you already can deduce the following This is saying when we Integrate either of the following, we get the same result: As we are integrating to find the Area, you can see for any 2 points, the area will be the same for either graph… Therefore you can use either x or –x in the Integral However, you cannot find ln of a negative, just use the positive value instead! 1 𝑥 = ln⁡|𝑥|+𝐶 𝒚= 𝟏 −𝒙 𝑓 𝑥 = 1 𝑥 𝑓 𝑥 = 1 −𝑥 6A

Integration 6A 2𝑐𝑜𝑠𝑥+ 3 𝑥 − 𝑥 𝑑𝑥 As the terms are separate you can integrate them separately 2𝑐𝑜𝑠𝑥+ 3 𝑥 − 𝑥 𝑑𝑥 You need to be able to integrate standard functions Find the following integral: 2𝑐𝑜𝑠𝑥 3 𝑥 𝑥 =2𝑠𝑖𝑛𝑥 =3ln⁡|𝑥| 2𝑐𝑜𝑠𝑥+ 3 𝑥 − 𝑥 𝑑𝑥 Rewrite this term as a power = 𝑥 1 2 = 𝑥 3 2 3 2 = 2 3 𝑥 3 2 =2𝑠𝑖𝑛𝑥+3 ln 𝑥 − 2 3 𝑥 3 2 +𝐶 Remember the + C! 6A

Integration 6A 𝑐𝑜𝑠𝑥 𝑠𝑖 𝑛 2 𝑥 −2 𝑒 𝑥 𝑑𝑥 As the terms are separate you can integrate them separately 𝑐𝑜𝑠𝑥 𝑠𝑖 𝑛 2 𝑥 −2 𝑒 𝑥 𝑑𝑥 You need to be able to integrate standard functions Find the following integral: 𝑐𝑜𝑠𝑥 𝑠𝑖 𝑛 2 𝑥 2 𝑒 𝑥 𝑐𝑜𝑠𝑥 𝑠𝑖 𝑛 2 𝑥 −2 𝑒 𝑥 𝑑𝑥 = 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 1 𝑠𝑖𝑛𝑥 =2 𝑒 𝑥 = 𝑐𝑜𝑡𝑥𝑐𝑜𝑠𝑒𝑐𝑥 Try to rewrite as an integral you ‘know’ =−𝑐𝑜𝑠𝑒𝑐𝑥 Remember the + C! =−𝑐𝑜𝑠𝑒𝑐𝑥−2 𝑒 𝑥 +𝐶 6A

Ex 6A C4

Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: 𝑦=sin⁡(2𝑥+3) Consider starting with sin(2x + 3) and the answer that would give 𝑑𝑦 𝑑𝑥 =2cos⁡(2𝑥+3) This is double what we are wanting to integrate  Therefore, we must ‘start’ with half the amount… cos 2𝑥+3 𝑑𝑥 cos 2𝑥+3 𝑑𝑥 Divide the original ‘guess’ by 2 = 1 2 sin 2𝑥+3 +𝐶 This is a VERY common method of integration – considering what we might start with that would differentiate to our answer… 6B

Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: 𝑦= 𝑒 4𝑥+1 Consider starting with e4x + 1 and the answer that would give 𝑑𝑦 𝑑𝑥 =4 𝑒 4𝑥+1 This is four times what we are wanting to integrate  Therefore, we must ‘start’ with a quarter of the amount… 𝑒 4𝑥+1 𝑑𝑥 𝑒 4𝑥+1 𝑑𝑥 Divide the original ‘guess’ by 4 = 1 4 𝑒 4𝑥+1 +𝐶 6B

Integration You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: 𝑦=𝑡𝑎𝑛3𝑥 Consider starting with tan3x and the answer that would give 𝑑𝑦 𝑑𝑥 =3𝑠𝑒 𝑐 2 3𝑥 This is three times what we are wanting to integrate  Therefore, we must ‘start’ with a third of the amount… 𝑠𝑒 𝑐 2 3𝑥 𝑑𝑥 𝑠𝑒 𝑐 2 3𝑥 𝑑𝑥 Divide the original ‘guess’ by 3 = 1 3 𝑡𝑎𝑛3𝑥+𝐶 6B

Integration 𝑓 ′ 𝑎𝑥+𝑏 = 1 𝑎 𝑓 𝑎𝑥+𝑏 +𝐶 6B cos 2𝑥+3 𝑑𝑥 𝑒 4𝑥+1 𝑑𝑥 𝑒 4𝑥+1 𝑑𝑥 𝑠𝑒 𝑐 2 3𝑥 𝑑𝑥 You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) These three answers illustrate a rule: = 1 2 sin 2𝑥+3 +𝐶 = 1 4 𝑒 4𝑥+1 +𝐶 = 1 3 𝑡𝑎𝑛3𝑥+𝐶 1) Integrate the function using what you know from C3 𝑓 ′ 𝑎𝑥+𝑏 = 1 𝑎 𝑓 𝑎𝑥+𝑏 +𝐶 2) Divide by the coefficient of x 3) Simplify if possible and add C 6B

𝑓 ′ 𝑎𝑥+𝑏 = 1 𝑎 𝑓 𝑎𝑥+𝑏 +𝐶 Integration 1 3𝑥+2 𝑑𝑥 1) Integrate the function using what you know from C3 You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: 2) Divide by the coefficient of x = 1 3 ln 3𝑥+2 +𝐶 3) Simplify if possible and add C 1 3𝑥+2 𝑑𝑥 6B

As this is 10 times what we want, we need to divide our ‘guess’ by 10 𝑓 ′ 𝑎𝑥+𝑏 = 𝑓 ′ 𝑎𝑥+𝑏 = 1 𝑎 𝑓 𝑎𝑥+𝑏 +𝐶 Integration 𝑦=(2𝑥+3 ) 5 You can integrate using the reverse of the Chain rule This technique will only work for linear transformations of functions such as f(ax + b) Find the following integral: Consider a function that would leave you with a power 4 and the same bracket 𝑑𝑦 𝑑𝑥 =5(2𝑥+3 ) 4 (2) Simplify after using the Chain rule 𝑑𝑦 𝑑𝑥 =10(2𝑥+3 ) 4 As this is 10 times what we want, we need to divide our ‘guess’ by 10 (2𝑥+3 ) 4 𝑑𝑥 (2𝑥+3 ) 4 𝑑𝑥 = 1 10 (2𝑥+3 ) 5 +𝐶 6B

Ex 6B C4

𝑠𝑖 𝑛 2 𝑥+𝑐𝑜 𝑠 2 𝑥≡1 Integration Divide by cos 𝑡𝑎 𝑛 2 𝑥+1≡𝑠𝑒 𝑐 2 𝑥 Subtract 1 𝑡𝑎 𝑛 2 𝑥≡𝑠𝑒 𝑐 2 𝑥−1 𝑡𝑎 𝑛 2 𝑥 𝑑𝑥 You can use Trigonometric Identities in Integration There are some trigonometric expressions you cannot integrate yet, but a way to do so is to write them in terms of ones you can integrate… Trigonometric Identities are invaluable in this! Find: Using the identity above, replace tan2x = (𝑠𝑒 𝑐 2 𝑥−1) 𝑑𝑥 Integrate each part separately 𝑠𝑒 𝑐 2 𝑥 1 =𝑡𝑎𝑛𝑥 =𝑥 =𝑡𝑎𝑛𝑥−𝑥+𝐶 𝑡𝑎 𝑛 2 𝑥 𝑑𝑥 6C

𝑐𝑜𝑠2𝑥=1−2𝑠𝑖 𝑛 2 𝑥 Integration Rearrange 2𝑠𝑖 𝑛 2 𝑥=1−𝑐𝑜𝑠2𝑥 Divide by 2 𝑠𝑖 𝑛 2 𝑥≡ 1 2 − 1 2 𝑐𝑜𝑠2𝑥 𝑠𝑖 𝑛 2 𝑥 𝑑𝑥 You can use Trigonometric Identities in Integration There are some trigonometric expressions you cannot integrate yet, but a way to do so is to write them in terms of ones you can integrate… Trigonometric Identities are invaluable in this! Find: Using the identity above, replace sin2x = 1 2 − 1 2 𝑐𝑜𝑠2𝑥 𝑑𝑥 Integrate each part separately 1 2 1 2 𝑐𝑜𝑠2𝑥 = 1 2 𝑥 = 1 4 𝑠𝑖𝑛2𝑥 𝑦=𝑠𝑖𝑛2𝑥 Use the ‘guessing’ method 𝑑𝑦 𝑑𝑥 =2𝑐𝑜𝑠2𝑥 = 1 2 𝑥− 1 4 𝑠𝑖𝑛2𝑥+𝐶 𝑠𝑖 𝑛 2 𝑥 𝑑𝑥 This is 4 times what we want so divide the ‘guess’ by 4 6C

Integration 6C 𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥 𝑑𝑥 s𝑖𝑛2𝑥=2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 Integration Double angle formula s𝑖𝑛4𝑥=2𝑠𝑖𝑛2𝑥𝑐𝑜𝑠2𝑥 Follow the pattern… s𝑖𝑛6𝑥=2𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥 Divide by 2 1 2 𝑠𝑖𝑛6𝑥=𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥 𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥 𝑑𝑥 You can use Trigonometric Identities in Integration There are some trigonometric expressions you cannot integrate yet, but a way to do so is to write them in terms of ones you can integrate… Trigonometric Identities are invaluable in this! Find: Replace with the above… 1 2 𝑠𝑖𝑛6𝑥 𝑑𝑥 This will give us the sin 6x when differentiating, but is negative and 12 times too big!  Make the guess negative and divide by 12! 𝑦=𝑐𝑜𝑠6𝑥 𝑑𝑦 𝑑𝑥 =−6𝑠𝑖𝑛6𝑥 1 2 𝑠𝑖𝑛6𝑥 𝑑𝑥 𝑠𝑖𝑛3𝑥𝑐𝑜𝑠3𝑥 𝑑𝑥 =− 1 12 𝑐𝑜𝑠6𝑥+𝐶 6C

𝑠𝑖 𝑛 2 𝑥+𝑐𝑜 𝑠 2 𝑥≡1 Integration Divide by cos 𝑡𝑎 𝑛 2 𝑥+1≡𝑠𝑒 𝑐 2 𝑥 Subtract 1 𝑡𝑎 𝑛 2 𝑥≡𝑠𝑒 𝑐 2 𝑥−1 (𝑠𝑒𝑐𝑥+𝑡𝑎𝑛𝑥 ) 2 𝑑𝑥 You can use Trigonometric Identities in Integration There are some trigonometric expressions you cannot integrate yet, but a way to do so is to write them in terms of ones you can integrate… Trigonometric Identities are invaluable in this! Find: Expand the bracket 𝑠𝑒 𝑐 2 𝑥+𝑡𝑎 𝑛 2 𝑥+2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 𝑑𝑥 Replace tan2x 𝑠𝑒 𝑐 2 𝑥+(𝑠𝑒 𝑐 2 𝑥−1)+2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 𝑑𝑥 Simplify 2𝑠𝑒 𝑐 2 𝑥−1+2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 𝑑𝑥 Integrate separately 2𝑠𝑒 𝑐 2 𝑥 1 2𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥 =2𝑡𝑎𝑛𝑥 =𝑥 =2𝑠𝑒𝑐𝑥 (𝑠𝑒𝑐𝑥+𝑡𝑎𝑛𝑥 ) 2 𝑑𝑥 =2𝑡𝑎𝑛𝑥−𝑥+2𝑠𝑒𝑐𝑥+𝐶 6C

What about …. 𝑠𝑖𝑛3𝑥𝑐𝑜𝑠2𝑥 𝑑𝑥

Ex 6C C4

Method ? ? Split 6𝑥−2 𝑥−3 𝑥+1 into partial fractions. Q If each factor of the denominator is linear, we can split like such (for constants 𝐴 and 𝐵): 6𝑥−2 𝑥−3 𝑥+1 ≡ 𝐴 𝑥−3 + 𝐵 𝑥+1 We don’t like fractions in equations, so we could simplify this to: 𝟔𝒙−𝟐≡𝑨 𝒙+𝟏 +𝑩 𝒙−𝟑 ? METHOD 1: Substitution ? We can easily eliminate either 𝐴 or 𝐵 by an appropriate choice of 𝑥: If 𝑥=−1: −8=−4𝐵 → 𝐵=2 If 𝑥=3: 16=4𝐴 → 𝐴=4 Therefore: 6𝑥−2 𝑥−3 𝑥+1 ≡ 4 𝑥−3 + 2 𝑥+1

Test Your Understanding C4 Jan 2011 Q3 ? 5 𝑥−1 3𝑥+2 ≡ 𝐴 𝑥−1 + 𝐵 3𝑥+2 5=𝐴 3𝑥+2 +𝐵 𝑥−1 Let 𝑥=1: 5=5𝐴 → 𝐴=1 Let 𝑥=− 2 3 : 5=− 5 3 𝐵 → 𝐵=−3 Therefore 5 𝑥−1 3𝑥+2 ≡ 1 𝑥−1 − 3 3𝑥+2 Notice we can move the “–” to the front of the fraction. Note that we don’t technically need this last line from the perspective of the mark scheme, but it’s good to just to be on the safe side

More than two fractions The principle is exactly the same if we have more than two linear factors in the denominator. Q Split 6 𝑥 2 +5𝑥−2 𝑥 𝑥−1 2𝑥+1 into partial fractions. ? 6 𝑥 2 +5𝑥−2 𝑥 𝑥−1 2𝑥+1 ≡ 𝐴 𝑥 + 𝐵 𝑥−1 + 𝐶 2𝑥+1 6 𝑥 2 +5𝑥−2≡𝐴 𝑥−1 2𝑥+1 +𝐵𝑥 2𝑥+1 +𝐶𝑥 𝑥−1 When 𝑥=0: −2=−𝐴 → 𝐴=2 When 𝑥=1: 9=3𝐵 → 𝐵=3 When 𝑥=− 1 2 : −3= 3 4 𝐶 → 𝐶=−4 So 6 𝑥 2 +5𝑥−2 𝑥 𝑥−1 2𝑥+1 ≡ 2 𝑥 + 3 𝑥−1 − 4 2𝑥+1 Bro Tip: While substitution is generally the easier method, I sometimes compare coefficients of just the 𝑥 2 term to avoid having to deal with fractions. No need to expand; we can see by observation that: 6=2𝐴+2𝐵+2𝐶 Then 𝐶 is easy to determine given we know 𝐴 and 𝐵.

Repeated linear factors Q Split 11 𝑥 2 +14𝑥+5 𝑥+1 2 2𝑥+1 into partial fractions. ? 11 𝑥 2 +14𝑥+5≡𝐴 𝑥+1 2𝑥+1 +𝐵 2𝑥+1 +𝐶 𝑥+1 2 When 𝑥=−1: 2=−𝐵 → 𝐵=−2 When 𝑥=− 1 2 : 3 4 = 1 4 𝐶 → 𝐶=3 At this point we could substitute something else (e.g. 𝑥=1) but it’s easier to equate 𝑥 2 terms. 11=2𝐴+𝐶 𝐴=4

Test Your Understanding C4 June 2011 Q1 ?

Dealing with Improper Fractions The ‘degree’ of a polynomial is the highest power, e.g. a quadratic has degree 2. An algebraic fraction is improper if the degree of the numerator is at least the degree of the denominator. 𝑥 3 − 𝑥 2 +3 𝑥 2 −𝑥 𝑥 2 −3 𝑥+2 𝑥+1 𝑥−1 ! To split an improper fraction into partial fractions, either: Divide algebraically first. Or introduce a whole term 𝐴+… and deal with identity immediately.

Dealing with Improper Fractions Q Split 3 𝑥 2 −3𝑥−2 𝑥−1 𝑥−2 into partial fractions. Method 1: Algebraic Division Method 2: Using One Identity (method not in your textbooks but in mark schemes) ? 3 𝑥 2 −3𝑥−2 𝑥−1 𝑥−2 ≡ 3 𝑥 2 −3𝑥−2 𝑥 2 −3𝑥+2 Dividing algebraically gives: 3+ 6𝑥−8 𝑥 2 −3𝑥+2 Turn numerator back: =3+ 6𝑥−8 𝑥−1 𝑥−2 Let 6𝑥−8 𝑥−1 𝑥−2 ≡ 𝐴 𝑥−1 + 𝐵 𝑥−2 𝐴=2 𝐵=4 So 𝟑 𝒙 𝟐 −𝟑𝒙−𝟐 𝒙−𝟏 𝒙−𝟐 ≡𝟑+ 𝟐 𝒙−𝟏 + 𝟒 𝒙−𝟐 ? Let: 3 𝑥 2 −3𝑥−2 𝑥−1 𝑥−2 ≡𝐴+ 𝐵 𝑥−1 + 𝐶 𝑥−2 3 𝑥 2 −3𝑥−2≡𝐴 𝑥−1 𝑥−2 +𝐵 𝑥−2 +𝐶 𝑥−1 If 𝑥=2: 4=𝐶 If 𝑥=1: −2=−𝐵 → 𝐵=2 Comparing coefficients of 𝑥 2 : 3=𝐴 opinion: I personally think the second method is easier. And mark schemes present it as “Method 1” – implying more standard!

Test Your Understanding C4 Jan 2013 Q3 ?

Integration 𝑥−5 (𝑥+1)(𝑥−2) = 𝐴 (𝑥+1) + 𝐵 (𝑥−2) You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: Write as two fractions and make the denominators equal 𝑥−5 (𝑥+1)(𝑥−2) = 𝐴(𝑥−2) (𝑥+1)(𝑥−2) + 𝐵(𝑥+1) (𝑥+1)(𝑥−2) Combine 𝑥−5 (𝑥+1)(𝑥−2) = 𝐴 𝑥−2 +𝐵(𝑥+1) (𝑥+1)(𝑥−2) The numerators must be equal 𝑥−5 =𝐴 𝑥−2 +𝐵(𝑥+1) Let x = 2 −3 =3𝐵 Calculate A and B by choosing appropriate x values −1 =𝐵 𝑥−5 (𝑥+1)(𝑥−2) Let x = -1 −6 =−3𝐴 2 =𝐴 2 (𝑥+1) − 1 (𝑥−2) 𝑥−5 (𝑥+1)(𝑥−2) = 𝐴 (𝑥+1) + 𝐵 (𝑥−2) Replace A and B from the start 𝑥−5 (𝑥+1)(𝑥−2) = 2 (𝑥+1) − 1 (𝑥−2) 6D

You can combine the natural logarithms as a division Integration 2 (𝑥+1) − 1 (𝑥−2) You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: Integrate separately 2 (𝑥+1) 1 (𝑥−2) =2ln⁡|𝑥+1| =ln⁡|𝑥−2| =ln⁡|(𝑥+1 ) 2 | =ln⁡|(𝑥+1 ) 2 |− ln 𝑥−2 +𝐶 𝑥−5 (𝑥+1)(𝑥−2) You can combine the natural logarithms as a division =ln⁡ 𝑥+1 2 𝑥−2 +𝐶 2 (𝑥+1) − 1 (𝑥−2) 6D

Integration 1 1) Divide the first term by the highest power You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: 9 𝑥 2 −4 9 𝑥 2 −3𝑥+2 9 𝑥 2 − 4 − 2) Multiply the answer by the whole expression you’re dividing by −3𝑥 + 6 3) Subtract to find the remainder 4) Remember to write the remainder as a fraction of the original expression 9 𝑥 2 −3𝑥+2 9 𝑥 2 −4 9 𝑥 2 −3𝑥+2 9 𝑥 2 −4 + −3𝑥+6 9 𝑥 2 −4 = 1 + 6−3𝑥 9 𝑥 2 −4 Looks tidier! = 1 6D

Integration 6D 9 𝑥 2 −3𝑥+2 9 𝑥 2 −4 + 6−3𝑥 9 𝑥 2 −4 = 1 + 6−3𝑥 9 𝑥 2 −4 We now need to write the remainder as partial fractions = 1 You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: 6−3𝑥 9 𝑥 2 −4 = 6−3𝑥 (3𝑥+2)(3𝑥−2) = 𝐴 (3𝑥+2) + 𝐵 (3𝑥−2) = 𝐴 3𝑥−2 +𝐵(3𝑥+2) (3𝑥−2)(3𝑥+2) 9 𝑥 2 −3𝑥+2 9 𝑥 2 −4 Set the numerators equal and solve for A and B 𝐴 3𝑥−2 +𝐵 3𝑥+2 =6−3𝑥 Let x = 2/3 4𝐵=4 𝐵=1 Let x = -2/3 −4𝐴=8 𝐴=−2 Write the final answer with the remainder broken apart! 1 + 6−3𝑥 9 𝑥 2 −4 = 1− 2 3𝑥+2 + 1 3𝑥−2 6D

Integration 6D 9 𝑥 2 −3𝑥+2 9 𝑥 2 −4 = 1− 2 3𝑥+2 + 1 3𝑥−2 = 1− 2 3𝑥+2 + 1 3𝑥−2 You can use partial fractions to integrate expressions This allows you to split a fraction up – it can sometimes be recombined after integration… Find: 1− 2 3𝑥+2 + 1 3𝑥−2 Integrate separately 1 2 3𝑥+2 1 3𝑥−2 = 2 1 3 ln⁡|3𝑥+2| = 1 3 ln⁡|3𝑥−2| =𝑥 9 𝑥 2 −3𝑥+2 9 𝑥 2 −4 = 1 3 ln⁡| 3𝑥+2 2 | =𝑥− 1 3 ln 3𝑥+2 2 + 1 3 ln 3𝑥−2 +𝐶 You can combine the natural logarithms (be careful, the negative goes on the bottom…) =𝑥+ 1 3 (3𝑥−2) (3𝑥+2) 2 +𝐶 6D

Ex 6D C4

Integration 6E You can Integrate by using standard patterns 1 2𝑥+3 You have seen how to integrate fractions of the form: Including using partial fractions where an expression can be factorised However, this method will not work for integrals of the form: Some expressions like this can by integrated by using the ‘standard patterns’ technique 1 2𝑥+3 1 𝑥 2 +1 6E

Integration Notice that the denominator would differentiate to become the numerator  This is a pattern we can use to figure out what the integral is… 2𝑥 𝑥 2 +1 𝑑𝑥 You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Remember… 𝑖𝑓 𝑦=ln⁡|𝑓 𝑥 | 𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥) 𝑓(𝑥) So imagine starting with ln|denominator| 𝑦=ln⁡| 𝑥 2 +1| 2𝑥 𝑥 2 +1 𝑑𝑥 In this case, we get straight to the answer! 𝑑𝑦 𝑑𝑥 = 2𝑥 𝑥 2 +1 2𝑥 𝑥 2 +1 𝑑𝑥 = ln 𝑥 2 +1 +𝐶 6E

Integration 6E 𝑖𝑓 𝑦=ln⁡|𝑓 𝑥 | 𝑑𝑦 𝑑𝑥 = 𝑓′(𝑥) 𝑓(𝑥) 𝑐𝑜𝑠𝑥 3+2𝑠𝑖𝑛𝑥 𝑑𝑥 𝑐𝑜𝑠𝑥 3+2𝑠𝑖𝑛𝑥 𝑑𝑥 You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Start by trying y = ln|denominator| 𝑦=ln⁡|3+2𝑠𝑖𝑛𝑥| Differentiate 𝑑𝑦 𝑑𝑥 = 2𝑐𝑜𝑠𝑥 3+2𝑠𝑖𝑛𝑥 This is double what we want so multiply the ‘guess’ by 1/2 𝑐𝑜𝑠𝑥 3+2𝑠𝑖𝑛𝑥 𝑑𝑥 𝑐𝑜𝑠𝑥 3+2𝑠𝑖𝑛𝑥 𝑑𝑥 = 1 2 ln 3+2𝑠𝑖𝑛𝑥 +𝐶 6E

Integration 6E 3𝑐𝑜𝑠𝑥𝑠𝑖 𝑛 2 𝑥 𝑑𝑥 In this case consider the power of sine.  If it has been differentiated, it must have been sin3x originally… 3𝑐𝑜𝑠𝑥𝑠𝑖 𝑛 2 𝑥 𝑑𝑥 You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: 𝑦=𝑠𝑖 𝑛 3 𝑥 Write as a cubed bracket 𝑦=(𝑠𝑖𝑛𝑥 ) 3 Differentiate using the chain rule 𝑑𝑦 𝑑𝑥 =3(𝑠𝑖𝑛𝑥 ) 2 (𝑐𝑜𝑠𝑥) Rewrite – this has given us exactly what we wanted! 3𝑐𝑜𝑠𝑥𝑠𝑖 𝑛 2 𝑥 𝑑𝑥 𝑑𝑦 𝑑𝑥 =3𝑐𝑜𝑠𝑥𝑠𝑖 𝑛 2 𝑥 3𝑐𝑜𝑠𝑥𝑠𝑖 𝑛 2 𝑥 𝑑𝑥 Don’t forget the + C! =𝑠𝑖 𝑛 3 𝑥+𝐶 6E

Integration Consider the power on the bracket  As it is a power 3, it must have been a power 4 before differentiation 𝑥( 𝑥 2 +5 ) 3 𝑑𝑥 You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: 𝑦=( 𝑥 2 +5 ) 4 Differentiate the bracket to the power 4 using the chain rule 𝑑𝑦 𝑑𝑥 =4( 𝑥 2 +5 ) 3 (2𝑥) Simplify 𝑑𝑦 𝑑𝑥 =8𝑥( 𝑥 2 +5 ) 3 𝑥( 𝑥 2 +5 ) 3 𝑑𝑥 This is 8 times too big so multiply the ‘guess’ by 1/8 𝑥( 𝑥 2 +5 ) 3 𝑑𝑥 Don’t forget to add C! = 1 8 ( 𝑥 2 +5 ) 4 +𝐶 6E

Integration 6E 𝑐𝑜𝑠𝑒 𝑐 2 𝑥 (2+𝑐𝑜𝑡𝑥 ) 3 𝑑𝑥 𝑐𝑜𝑠𝑒 𝑐 2 𝑥 (2+𝑐𝑜𝑡𝑥 ) 3 𝑑𝑥 You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Write using powers (𝑐𝑜𝑠𝑒 𝑐 2 𝑥)(2+𝑐𝑜𝑡𝑥 ) −3 𝑑𝑥 Imagine how we could end up with a -3 as a power... 𝑦=(2+𝑐𝑜𝑡𝑥 ) −2 Use the chain rule 𝑑𝑦 𝑑𝑥 =−2(2+𝑐𝑜𝑡𝑥 ) −3 (−𝑐𝑜𝑠𝑒 𝑐 2 𝑥) Rewrite 𝑑𝑦 𝑑𝑥 =2𝑐𝑜𝑠𝑒 𝑐 2 𝑥(2+𝑐𝑜𝑡𝑥 ) −3 𝑐𝑜𝑠𝑒 𝑐 2 𝑥 (2+𝑐𝑜𝑡𝑥 ) 3 𝑑𝑥 This is double what we want so multiply the ‘guess’ by 1/2 𝑑𝑦 𝑑𝑥 = 2𝑐𝑜𝑠𝑒 𝑐 2 𝑥 (2+𝑐𝑜𝑡𝑥 ) 3 𝑐𝑜𝑠𝑒 𝑐 2 𝑥 (2+𝑐𝑜𝑡𝑥 ) 3 𝑑𝑥 = 1 2 (2+𝑐𝑜𝑡𝑥 ) −2 +𝐶 6E

Integration 6E 5𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 4 𝑥 𝑑𝑥 5𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 4 𝑥 𝑑𝑥 You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Consider using a power 5 𝑦=𝑠𝑒 𝑐 5 𝑥 Write as a bracket to the power 5 𝑦=(𝑠𝑒𝑐𝑥 ) 5 Differentiate using the chain rule 𝑑𝑦 𝑑𝑥 =5(𝑠𝑒𝑐𝑥 ) 4 (𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥) We have an extra secx 𝑑𝑦 𝑑𝑥 =5𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 5 𝑥 5𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 4 𝑥 𝑑𝑥 HOWEVER: We cannot just add this to our ‘guess’ as before, as the differentiation will need to be performed using the product rule from C3, rather than the Chain rule! We need to find another way! 6E

Integration 6E 5𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 4 𝑥 𝑑𝑥 5𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 4 𝑥 𝑑𝑥 You can Integrate by using standard patterns You have seen how to integrate fractions of the form: Some expressions can by integrated by using the ‘standard patterns’ technique Find: Consider using a power 4 𝑦=𝑠𝑒 𝑐 4 𝑥 Write as a bracket to the power 4 𝑦=(𝑠𝑒𝑐𝑥 ) 4 Differentiate using the chain rule 𝑑𝑦 𝑑𝑥 =4(𝑠𝑒𝑐𝑥 ) 3 (𝑠𝑒𝑐𝑥𝑡𝑎𝑛𝑥) 𝑑𝑦 𝑑𝑥 =4𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 4 𝑥 5𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 4 𝑥 𝑑𝑥 This is what we want, but 4/5 of the amount  Multiply by the guess by 5/4 5𝑡𝑎𝑛𝑥𝑠𝑒 𝑐 4 𝑥 𝑑𝑥 𝑦= 5 4 𝑠𝑒 𝑐 4 𝑥+𝐶 6E

Ex 6E C4

Integration To integrate this, you need to replace the x terms with equivalent u terms, and replace the dx with an equivalent du 𝑥 2𝑥+5 𝑑𝑥 It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: 𝑢=2𝑥+5 Differentiate Rearrange to find x 𝑑𝑢 𝑑𝑥 =2 𝑢−5=2𝑥 𝑢=2𝑥+5 Rearrange to get dx 1 2 (𝑢−5)=𝑥 𝑑𝑢 2 =𝑑𝑥 𝑥 2𝑥+5 𝑑𝑥 𝑥 2𝑥+5 𝑑𝑥 Replace each ‘x’ term with an equivalent ‘u’ term 1 2 (𝑢−5) 𝑑𝑢 2 𝑢 Rearrange – you should leave ‘du’ at the end 1 4 𝑢−5 𝑢 𝑑𝑢 Combine terms including the square root, changed to a power ‘1/2’ 1 4 𝑢 3 2 − 5 4 𝑢 1 2 𝑑𝑢 6F

Integration 1 4 𝑢 3 2 − 5 4 𝑢 1 2 𝑑𝑢 It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: Differentiate terms separately = 1 4 𝑢 5 2 5 2 − 5 4 𝑢 3 2 3 2 + 𝐶 Flip the dividing fractions = 1 4 2 5 𝑢 5 2 − 5 4 2 3 𝑢 3 2 𝑢=2𝑥+5 + 𝐶 Calculate the fraction parts 𝑥 2𝑥+5 𝑑𝑥 = 1 5 𝑢 5 2 − 5 6 𝑢 3 2 + 𝐶 Finally, replace with u with its equivalent from the start! = 1 5 (2𝑥+5) 5 2 − 5 6 (2𝑥+5) 3 2 + 𝐶 6F

Integration 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥(1+𝑠𝑖𝑛𝑥 ) 3 𝑑𝑥 It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use the substitution: To find: 𝑢=𝑠𝑖𝑛𝑥+1 Differentiate Rearrange to find Sinx 𝑑𝑢 𝑑𝑥 =𝑐𝑜𝑠𝑥 𝑢−1=𝑠𝑖𝑛𝑥 Rearrange to get dx 𝑑𝑢 𝑐𝑜𝑠𝑥 =𝑑𝑥 𝑢=𝑠𝑖𝑛𝑥+1 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥(1+𝑠𝑖𝑛𝑥 ) 3 𝑑𝑥 Replace each ‘x’ term with an equivalent ‘u’ term 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥(1+𝑠𝑖𝑛𝑥 ) 3 𝑑𝑥 𝑐𝑜𝑠𝑥 𝑑𝑢 𝑐𝑜𝑠𝑥 (𝑢−1) 𝑢 3 Cancel the Cosx terms 𝑢−1 𝑢 3 𝑑𝑢 Multiply out 𝑢 4 − 𝑢 3 𝑑𝑢 Integrate = 1 5 𝑢 5 − 1 4 𝑢 4 +𝐶 Replace u with x terms again! = 1 5 (𝑠𝑖𝑛𝑥+1) 5 − 1 4 (𝑠𝑖𝑛𝑥+1) 4 +𝐶 6F

Integration 0 2 𝑥(𝑥+1 ) 3 𝑑𝑥 It is sometimes possible to simplify an integral by changing the variable. This is known as integration by substitution. Use integration by substitution to find: Sometimes you will have to decide on a substitution yourself. In this case, the bracket would be hardest to integrate so it makes sense to use the substitution: 𝑢=𝑥+1 Rearrange to find x Differentiate 𝑑𝑢 𝑑𝑥 =1 𝑢−1=𝑥 Rearrange to get dx 𝑑𝑢=𝑑𝑥 𝑢=𝑥+1 You also need to recalculate limits in terms of u 0 2 𝑥(𝑥+1 ) 3 𝑑𝑥 𝑥=2, 𝑢=3 𝑥=0, 𝑢=1 0 2 𝑥(𝑥+1 ) 3 𝑑𝑥 Replace x limits with u limits and the x terms with u terms 1 3 (𝑢−1) 𝑢 3 𝑑𝑢 Multiply out bracket 1 3 𝑢 4 − 𝑢 3 𝑑𝑢 Integrate = 𝑢 5 5 − 𝑢 4 4 1 3 𝑢=𝑥+1 An alternative method is to replace the ‘u’ terms with x terms at the end and then just use the original ‘x’ limits – either way is fine! Sub in limits and calculate = 3 5 5 − 3 4 4 − 1 5 5 − 1 4 4 6F =28.4

Ex 6F C4

Integration 𝑡𝑎𝑛𝑥 𝑑𝑥= ln 𝑠𝑒𝑐𝑥 +𝐶 𝑐𝑜𝑡𝑥 𝑑𝑥= ln 𝑠𝑖𝑛𝑥 +𝐶 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 Integration You can use integration by parts to integrate some expressions You may need the following Integrals, which you are given in the formula booklet… 𝑡𝑎𝑛𝑥 𝑑𝑥= ln 𝑠𝑒𝑐𝑥 +𝐶 𝑐𝑜𝑡𝑥 𝑑𝑥= ln 𝑠𝑖𝑛𝑥 +𝐶 𝑐𝑜𝑠𝑒𝑐𝑥 𝑑𝑥= −ln 𝑐𝑜𝑠𝑒𝑐𝑥+𝑐𝑜𝑡𝑥 +𝐶 𝑠𝑒𝑐𝑥 𝑑𝑥= ln 𝑠𝑒𝑐𝑥+𝑡𝑎𝑛𝑥 +𝐶 6G

You can use integration by parts to integrate some expressions In C3 you met the following: (the product rule) You can use integration by parts to integrate some expressions 𝑑 𝑑𝑥 𝑢𝑣 = 𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥 This is the differential of two functions multiplied together  You could think of it as: Rearrange by subtracting vdu/dx 𝑢 𝑑𝑣 𝑑𝑥 = 𝑑 𝑑𝑥 𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑑(𝑢𝑣) 𝑑𝑥 Integrate each term with respect to x 𝑢 𝑑𝑣 𝑑𝑥 = 𝑑 𝑑𝑥 𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 The middle term is just a differential  Integrating a differential cancels them both out! 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 This is the formula used for Integration by parts!  You get given this in the booklet The other terms do not cancel as only part of them are differentiated… As a general rule, it is easiest to let u = anything of the form xn. The exception is when there is a lnx term, in which case this should be used as u 6G

𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 Integration Unlike when using the product rule, we now have one function to differentiate, and one to integrate… 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 You can use integration by parts to integrate some expressions Find: You can recognise that Integration by parts is needed as we have two functions multiplied together… 𝑢=𝑥 𝑣=𝑠𝑖𝑛𝑥 𝑑𝑢 𝑑𝑥 =1 𝑑𝑣 𝑑𝑥 =𝑐𝑜𝑠𝑥 𝑥𝑐𝑜𝑠𝑥 𝑑𝑥 Differentiate Integrate Now replace the relevant parts to find the integral… − (𝑠𝑖𝑛𝑥)(1) =(𝑥)(𝑠𝑖𝑛𝑥) The integral here is simpler! =𝑥𝑠𝑖𝑛𝑥 − (−𝑐𝑜𝑠𝑥) Be careful with negatives here! =𝑥𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 + 𝐶 As a general rule, it is easiest to let u = anything of the form xn. The exception is when there is a lnx term, in which case this should be used as u 6G

Integration 6G 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑥 2 𝑙𝑛𝑥 𝑑𝑥 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 Integration 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑣= 𝑥 3 3 You can use integration by parts to integrate some expressions Find: You can recognise that Integration by parts is needed as we have two functions multiplied together… 𝑢=𝑙𝑛𝑥 𝑑𝑢 𝑑𝑥 = 1 𝑥 𝑑𝑣 𝑑𝑥 = 𝑥 2 𝑥 2 𝑙𝑛𝑥 𝑑𝑥 Differentiate Integrate Now replace the relevant parts to find the integral… =(𝑙𝑛𝑥) 𝑥 3 3 − 𝑥 3 3 1 𝑥 Simplify terms = 𝑥 3 3 𝑙𝑛𝑥 − 1 3 𝑥 2 Let u be lnx! Integrate the second part = 𝑥 3 3 𝑙𝑛𝑥 − 𝑥 3 9 + 𝐶 6G

Integration 6G 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑥 2 𝑒 𝑥 𝑑𝑥 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 Integration 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 𝑢= 𝑥 2 𝑣= 𝑒 𝑥 You can use integration by parts to integrate some expressions Find: You can recognise that Integration by parts is needed as we have two functions multiplied together… Sometimes you will have to use the process twice! This happens if the new integral still has two functions multiplied together… 𝑑𝑢 𝑑𝑥 =2𝑥 𝑑𝑣 𝑑𝑥 = 𝑒 𝑥 Differentiate Integrate Now replace the relevant parts to find the integral… 𝑥 2 𝑒 𝑥 𝑑𝑥 − ( 𝑒 𝑥 )(2𝑥) =( 𝑥 2 )( 𝑒 𝑥 ) − 2𝑥 𝑒 𝑥 = 𝑥 2 𝑒 𝑥 𝑢=2𝑥 𝑣= 𝑒 𝑥 𝑑𝑢 𝑑𝑥 =2 𝑑𝑣 𝑑𝑥 = 𝑒 𝑥 Differentiate Integrate − 2𝑥 𝑒 𝑥 − 2 𝑒 𝑥 = 𝑥 2 𝑒 𝑥 Work out the square bracket which is the second integration by parts = 𝑥 2 𝑒 𝑥 − 2𝑥 𝑒 𝑥 −2 𝑒 𝑥 Careful with negatives!! = 𝑥 2 𝑒 𝑥 − 2𝑥 𝑒 𝑥 + 2 𝑒 𝑥 + 𝐶 6G

You can use integration by parts to integrate some expressions 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 Integration When integrating lnx, you MUST think of it as ‘lnx times 1, and use lnx as ‘u’ and 1 as ‘dv/dx’ 𝑢 𝑑𝑣 𝑑𝑥 =𝑢𝑣 − 𝑣 𝑑𝑢 𝑑𝑥 You can use integration by parts to integrate some expressions Evaluate: Leave your answer in terms of natural logarithms… You will be asked to leave exact answers a lot so make sure you know your log laws!! 𝑢=𝑙𝑛𝑥 𝑣=𝑥 𝑑𝑢 𝑑𝑥 = 1 𝑥 𝑑𝑣 𝑑𝑥 =1 1 2 𝑙𝑛𝑥 𝑑𝑥 Differentiate Integrate Now replace the relevant parts to find the integral… − (𝑥) 1 𝑥 =(𝑙𝑛𝑥)(𝑥) Simplify terms − 1 =𝑥𝑙𝑛𝑥 Integrate and use a square bracket with limits = 𝑥𝑙𝑛𝑥−𝑥 1 2 Sub in the limits = 2𝑙𝑛2−2 − (1𝑙𝑛1−1) Calculate and leave in terms of ln2 =2𝑙𝑛2−1 6G

Integrating ln 𝑥 and definite integration Q Find ln 𝑥 𝑑𝑥 , leaving your answer in terms of natural logarithms. 𝑢= ln 𝑥 𝑑𝑣 𝑑𝑥 =1 𝑑𝑢 𝑑𝑥 = 1 𝑥 𝑣=𝑥 ln 𝑥 𝑑𝑥 =𝑥 ln 𝑥 − 1 𝑑𝑥 =𝑥 ln 𝑥 −𝑥+𝐶 ? Q Find 1 2 ln 𝑥 𝑑𝑥 , leaving your answer in terms of natural logarithms. 1 2 ln 𝑥 𝑑𝑥 = 𝑥 ln 𝑥 −𝑥 1 2 =𝟐 𝐥𝐧 𝟐 −𝟏 If we were doing it from scratch: 𝑢= ln 𝑥 𝑑𝑣 𝑑𝑥 =1 𝑑𝑢 𝑑𝑥 = 1 𝑥 𝑣=𝑥 1 2 ln 𝑥 𝑑𝑥 = 𝒙 𝒍𝒏 𝒙 𝟏 𝟐 − 1 2 1 𝑑𝑥 =2 ln 2 −1 ln 1 − 𝑥 1 2 =2 ln 2 −(2−1) =2 ln 2 −1 ? In general: 𝑎 𝑏 𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥 = 𝑢𝑣 𝑎 𝑏 − 𝑎 𝑏 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥

Ex 6G C4

Overview Integration by ‘reverse chain’ rule (We imagine what would have differentiated to get the expression.) Integration by standard result (There’s certain expressions you’re expected to know straight off.) sec 2 𝑥 𝑑𝑥= tan 𝑥 +𝐶 sin 3 𝑥 cos 𝑥 𝑑𝑥= 1 4 sin 4 𝑥 +𝐶 Integration by parts (Allows us to integrate a product, just as the product rule allowed us to differentiate one) Integration by substitution (We make a substitution to hopefully make the expression easier to integrate) 𝑥 cos 𝑥 𝑑𝑥 𝑢=𝑥 𝑑𝑣 𝑑𝑥 = cos 𝑥 𝑑𝑢 𝑑𝑥 =1 𝑣= sin 𝑥 𝑥 cos 𝑥 𝑑𝑥=𝑥 sin 𝑥 − sin 𝑥 𝑑𝑥 =𝑥 sin 𝑥 + cos 𝑥 𝑥 2𝑥+5 𝑑𝑥 Let 𝑢=2𝑥+5 → 𝑥= 𝑢−5 2 𝑑𝑢 𝑑𝑥 =2 → 𝑑𝑥= 1 2 𝑢 𝑥 2𝑥+5 𝑑𝑥= 𝑢−5 2 1 2 𝑑𝑢=…

Test Your Understanding Q Find 0 𝜋 2 𝑥 sin 𝑥 𝑑𝑥 ? 𝑢=𝑥 𝑑𝑣 𝑑𝑥 = sin 𝑥 𝑑𝑢 𝑑𝑥 =1 𝑣=− cos 𝑥 𝑥 sin 𝑥 𝑑𝑥 =−𝑥 cos 𝑥 − − cos 𝑥 𝑑𝑥 =−𝑥 cos 𝑥 + cos 𝑥 𝑑𝑥 =−𝑥 cos 𝑥 + sin 𝑥 𝑑𝑥 ∴ 0 𝜋 2 𝑥 sin 𝑥 𝑑𝑥 = −𝑥 cos 𝑥 + sin 𝑥 0 𝜋 2 = − 𝜋 2 cos 𝜋 2 + sin 𝜋 2 − 0+ sin 0 =1

Summary of Functions ? ? ? ? ? ? ? ? ? 𝒇(𝒙) How to deal with it 𝒇 𝒙 𝒅𝒙 (+constant) Formula booklet? 𝒔𝒊𝒏 𝒙 Standard result − cos 𝑥 No 𝒄𝒐𝒔 𝒙 sin 𝑥 𝒕𝒂𝒏 𝒙 In formula booklet, but use sin 𝑥 cos 𝑥 𝑑𝑥 which is of the form 𝑘 𝑓 ′ 𝑥 𝑓 𝑥 𝑑𝑥 ln sec 𝑥 Yes 𝒔𝒊 𝒏 𝟐 𝒙 For both sin 2 𝑥 and cos 2 𝑥 use identities for cos 2𝑥 cos 2𝑥 =1−2 sin 2 𝑥 sin 2 𝑥 = 1 2 − 1 2 cos 2𝑥 1 2 𝑥− 1 4 sin 2𝑥 𝒄𝒐 𝒔 𝟐 𝒙 cos 2𝑥 =2 cos 2 𝑥 −1 cos 2 𝑥 = 1 2 + 1 2 cos 2𝑥 1 2 𝑥+ 1 4 sin 2𝑥 𝒕𝒂 𝒏 𝟐 𝒙 1+ tan 2 𝑥 ≡ sec 2 𝑥 tan 2 𝑥 ≡ sec 2 𝑥 −1 tan 𝑥 −𝑥 𝒄𝒐𝒔𝒆𝒄 𝒙 Would use substitution 𝑢=𝑐𝑜𝑠𝑒𝑐 𝑥+ cot 𝑥 , but too hard for exam. −ln 𝑐𝑜𝑠𝑒𝑐 𝑥+ cot 𝑥 𝒔𝒆𝒄 𝒙 Would use substitution 𝑢= sec 𝑥 + tan 𝑥 , but too hard for exam. ln sec 𝑥 + tan 𝑥 𝒄𝒐𝒕 𝒙 cos 𝑥 sin 𝑥 𝑑𝑥 which is of the form 𝑓 ′ 𝑥 𝑓 𝑥 𝑑𝑥 ln 𝑠𝑖𝑛 𝑥 ? ? ? ? ? ? ? ? ?

Summary of Functions ? ? ? ? ? ? ? ? 𝒇(𝒙) How to deal with it 𝒇 𝒙 𝒅𝒙 (+constant) Formula booklet? 𝒄𝒐𝒔𝒆 𝒄 𝟐 𝒙 By observation. − 𝐜𝐨𝐭 𝒙 No! 𝒔𝒆 𝒄 𝟐 𝒙 tan 𝑥 Yes (but memorise) 𝒄𝒐 𝒕 𝟐 𝒙 1+ cot 2 𝑥 ≡𝑐𝑜𝑠𝑒 𝑐 2 𝑥 − cot 𝑥 −𝑥 No 𝒔𝒊𝒏 𝟐𝒙 𝒄𝒐𝒔 𝟐𝒙 For any product of sin and cos with same coefficient of 𝑥, use double angle. sin 2𝑥 cos 2𝑥 ≡ 1 2 sin 4𝑥 − 1 8 cos 4𝑥 𝟏 𝒙   ln 𝑥 𝐥𝐧 𝒙 Use IBP, where 𝑢= ln 𝑥 , 𝑑𝑣 𝑑𝑥 = ln 𝑥 𝑥 ln 𝑥 −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 ? ? ? ? ? ? ? ?

Summary of Functions ? ? ? ? ? 𝟒𝒙 𝒙 𝟐 +𝟏 𝒇(𝒙) How to deal with it 𝒇 𝒙 𝒅𝒙 (+constant) 𝟒𝒙 𝒙 𝟐 +𝟏 Reverse chain rule. Of form 𝑘 𝑓 ′ 𝑥 𝑓 𝑥 2 ln 𝑥 2 +1 𝒙 𝒙 𝟐 +𝟏 𝟐 Power around denominator so NOT of form 𝑘 𝑓 ′ 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 𝒆 𝟐𝒙+𝟏 𝟏 𝟏−𝟑𝒙 For any function where ‘inner function’ is linear expression, divide by coefficient of 𝑥 1 2 𝑒 2𝑥+1 − 1 3 ln 1−3𝑥 𝒙 𝟐𝒙+𝟏 Use sensible substitution. 𝑢=2𝑥+1 or even better, 𝑢 2 =2𝑥+1. 1 15 2𝑥+1 3 2 3𝑥−1 𝐬𝐢𝐧 𝟓 𝒙 𝒄𝒐𝒔 𝒙 Reverse chain rule. 1 6 sin 6 𝑥 ? ? ? ? ?