Integration by Substitution For some integrations involving a complicated expression, we can make a substitution to turn it into an equivalent integration that is simpler. We wouldn’t be able to use ‘reverse chain rule’ on the following: Q Use the substitution 𝑢=2𝑥+5 to find 𝑥 2𝑥+5 𝑑𝑥 The aim is to completely remove any reference to 𝑥, and replace it with 𝑢. We’ll have to work out 𝑥 and 𝑑𝑥 so that we can replace them. STEP 1: Using substitution, work out 𝑥 and 𝑑𝑥 (or variant) ? 𝑑𝑢 𝑑𝑥 =2 → 𝑑𝑥= 1 2 𝑑𝑢 Bro Tip: Be careful about ensuring you reciprocate when rearranging. 𝑥= 𝑢−5 2 𝑥 2𝑥+5 𝑑𝑥= 𝑢−5 2 𝑢 1 2 𝑑𝑢 = 1 4 𝑢 𝑢−5 𝑑𝑢 = 1 4 𝑢 3 2 −5 𝑢 1 2 = 1 4 2 5 𝑢 5 2 − 10 3 𝑢 3 2 +𝐶 = 2𝑥+5 5 2 10 − 5 2𝑥+5 3 2 6 +𝐶 ? STEP 2: Substitute these into expression. Bro Tip: If you have a constant factor, factor it out of the integral. STEP 3: Integrate simplified expression. ? STEP 4: Write answer in terms of 𝑥. ?

How can we tell what substitution to use? In Edexcel you will usually be given the substitution! However in some other exam boards, and in STEP, you often aren’t. There’s no hard and fast rule, but it’s often helpful to replace to replace expressions inside roots, powers or the denominator of a fraction. Sensible substitution: cos 𝑥 1+ sin 𝑥 𝑑𝑥 ? 𝒖=𝟏+ 𝐬𝐢𝐧 𝒙 6𝑥 𝑒 𝑥 2 𝑑𝑥 ? But this can be integrated by inspection. 𝒖= 𝒙 𝟐 𝑥 𝑒 𝑥 1+𝑥 𝑑𝑥 ? 𝒖=𝟏+𝒙 𝑒 1−𝑥 1+𝑥 𝑑𝑥 𝒖= 𝟏−𝒙 𝟏+𝒙 ?

Another Example Q Use the substitution 𝑢= sin 𝑥 +1 to find cos 𝑥 sin 𝑥 1+ sin 𝑥 3 𝑑𝑥 STEP 1: Using substitution, work out 𝑥 and 𝑑𝑥 (or variant) ? 𝑢= sin 𝑥 +1 𝑑𝑢 𝑑𝑥 = cos 𝑥 → 𝑑𝑢= cos 𝑥 𝑑𝑥 sin 𝑥 =𝑢−1 Notice this time we didn’t find 𝒙 or 𝒅𝒙. We could, but then 𝒙= 𝐚𝐫𝐜𝐬𝐢𝐧 (𝒖−𝟏) , and it would be slightly awkward simplifying the expression (although is still very much a valid method!) STEP 2: Substitute these into expression. ? = 𝑢−1 𝑢 3 𝑑𝑢 = 𝑢 4 − 𝑢 3 𝑑𝑢 = 1 5 𝑢 5 − 1 4 𝑢 4 +𝐶 = 1 5 sin 𝑥 +1 5 − 1 4 sin 𝑥 +1 4 +𝐶 STEP 3: Integrate simplified expression. ? STEP 4: Write answer in terms of 𝑥. ?

Using substitutions involving implicit differentiation When a root is involved, it makes thing much tidier if we use 𝑢 2 =… Q Use the substitution 𝑢 2 =2𝑥+5 to find 𝑥 2𝑥+5 𝑑𝑥 STEP 1: Using substitution, work out 𝑥 and 𝑑𝑥 (or variant) ? 2𝑢 𝑑𝑢 𝑑𝑥 =2 → 𝑑𝑥=𝑢 𝑑𝑢 𝑥= 𝑢 2 −5 2 𝑥 2𝑥+5 𝑑𝑥= 𝑢 2 −5 2 𝑢×𝑢𝑑𝑢 = 1 2 𝑢 4 − 5 2 𝑢 2 𝑑𝑢 = 1 10 𝑢 5 − 5 6 𝑢 3 +𝐶 = 2𝑥+5 5 2 10 − 5 2𝑥+5 3 2 6 +𝐶 ? STEP 2: Substitute these into expression. STEP 3: Integrate simplified expression. ? STEP 4: Write answer in terms of 𝑥. ? This was marginally less tedious than when we used 𝑢=2𝑥+5, as we didn’t have fractional powers to deal with.

Test Your Understanding Edexcel C4 Jan 2012 Q6c Hint: You might want to use your double angle formula first. ? 𝑑𝑢 𝑑𝑥 =− sin 𝑥 → 𝑑𝑥=− 1 sin 𝑥 𝑑𝑢 cos 𝑥 =𝑢−1 (As before 𝑥= arccos 𝑢−1 is going to be messy) 2 sin 2𝑥 1+ cos 𝑥 𝑑𝑥 = 4 sin 𝑥 cos 𝑥 1+ cos 𝑥 𝑑𝑥 = − 4 sin 𝑥 𝑢−1 𝑢 1 sin 𝑥 𝑑𝑢 =−4 𝑢−1 𝑢 𝑑𝑢 =−4 1− 1 𝑢 𝑑𝑢 =−4 𝑢− ln 𝑢 +𝑐=−4 1+ cos 𝑥 − ln cos 𝑥 +1 +𝑐=…

Definite Integration ? ? ? ? Now consider: Q Calculate 0 𝜋 2 cos 𝑥 1+ sin 𝑥 𝑑𝑥 ? Use substitution: 𝒖=𝟏+ 𝐬𝐢𝐧 𝒙 𝑑𝑢 𝑑𝑥 = cos 𝑥 → 𝑑𝑢= cos 𝑥 𝑑𝑥 sin 𝑥 =𝑢−1 Now because we’ve changed from 𝒙 to 𝒖, we have to work out what values of 𝒖 would have given those limits for 𝒙: When 𝑥= 𝜋 2 , 𝑢=2 When 𝑥=0, 𝑢=1 0 𝜋 2 cos 𝑥 1+ sin 𝑥 𝑑𝑥= 1 2 𝑢 1 2 𝑑𝑢 = 2 3 𝑢 3 2 1 2 = 2 3 2 2 −1 ? ? ?

Test Your Understanding Edexcel C4 June 2011 Q4 ? 𝒅𝒖 𝒅𝒙 =𝟐𝒙 → 𝒅𝒖=𝟐𝒙 𝒅𝒙 𝒙 𝟐 =𝒖−𝟐 When 𝒙= 𝟐 , 𝒖= 𝟐 𝟐 +𝟐=𝟒 When 𝒙=𝟎, 𝒖= 𝟎 𝟐 +𝟐=𝟐 𝟎 𝟐 𝒙 𝟐 𝐥𝐧 𝒙 𝟐 +𝟐 𝒙 𝒅𝒙= 𝟐 𝟒 𝟏 𝟐 (𝒖−𝟐) 𝐥𝐧 𝒖 𝒅𝒖

Exercise 6F ? ? ? ? ? ? ? ? ? Use the given substitution to integrate. Use an appropriate substitution. 𝑥 1+𝑥 𝑑𝑥 ; 𝑢=1+𝑥 = 𝟐 𝟓 𝟏+𝒙 𝟓 𝟐 − 𝟐 𝟑 𝟏+𝒙 𝟑 𝟐 +𝑪 1+ sin 𝑥 cos 𝑥 𝑑𝑥;𝑢= sin 𝑥 =− 𝐥𝐧 𝟏− 𝐬𝐢𝐧 𝒙 +𝑪 sin 3 𝑥 𝑑𝑥 ; 𝑢= cos 𝑥 = 𝟏 𝟑 𝐜𝐨𝐬 𝟑 𝒙 − 𝐜𝐨𝐬 𝒙 +𝑪 𝑥 2+𝑥 𝑑𝑥 ; 𝑢 2 =2+𝑥 = 𝟐 𝟓 𝟐+𝒙 𝟓 𝟐 − 𝟒 𝟑 𝟐+𝒙 𝟑 𝟐 +𝑪 sec 2 𝑥 tan 𝑥 1+ tan 𝑥 𝑑𝑥 ; 𝑢 2 =1+ tan 𝑥 = 𝟐 𝟓 𝟏+ 𝐭𝐚𝐧 𝒙 𝟓 𝟐 − 𝟐 𝟑 𝟏+ 𝐭𝐚𝐧 𝒙 𝟑 𝟐 +𝑪 sec 4 𝑥 𝑑𝑥 ; 𝑢= tan 𝑥 = 𝐭𝐚𝐧 𝒙 + 𝟏 𝟑 𝐭𝐚𝐧 𝟑 𝒙 +𝑪 1 a 0 5 𝑥 𝑥+4 𝑑𝑥 = 𝟓𝟎𝟔 𝟏𝟑 2 5 1 1+ 𝑥−1 𝑑𝑥 =𝟐+𝟐 𝐥𝐧 𝟐 𝟑 0 1 𝑥 2+𝑥 3 𝑑𝑥 =𝟗.𝟕 ? 3 a ? c c ? ? e ? e ? 2 a ? c ? e ?

Integration by Parts Proof ? 𝑥 cos 𝑥 𝑑𝑥 =? ! To integrate by parts: Just as the Product Rule was used to differentiate the product of two expressions, we can often use ‘Integration by Parts’ to integrate a product. ! To integrate by parts: 𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥=𝑢𝑣− 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥 Proof ? (not needed for exam) The Product Rule: 𝑑 𝑑𝑥 𝑢𝑣 =𝑣 𝑑𝑢 𝑑𝑥 +𝑢 𝑑𝑣 𝑑𝑥 On the right-hand-side, both 𝑣 𝑑𝑢 𝑑𝑥 and 𝑢 𝑑𝑣 𝑑𝑥 are the product of two expressions. So if we made either the subject, we could use 𝑢 𝑑𝑣 𝑑𝑥 say to represent 𝑥 cos 𝑥 in the example. Rearranging: 𝑢 𝑑𝑣 𝑑𝑥 = 𝑑 𝑑𝑥 𝑢𝑣 −𝑣 𝑑𝑢 𝑑𝑥 Integrating both sides with respect to 𝑥, we get the desired formula.

Integration by Parts ? ? ? 𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥=𝑢𝑣− 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥 𝑥 cos 𝑥 𝑑𝑥 =? 𝑢 𝑑𝑣 𝑑𝑥 𝑑𝑥=𝑢𝑣− 𝑣 𝑑𝑢 𝑑𝑥 𝑑𝑥 𝑥 cos 𝑥 𝑑𝑥 =? ? 𝑢=𝑥 𝑑𝑣 𝑑𝑥 = cos 𝑥 𝑑𝑢 𝑑𝑥 =1 𝑣= sin 𝑥 𝑥 cos 𝑥 𝑑𝑥 =𝑥 sin 𝑥 − 1 sin 𝑥 𝑑𝑥 =𝑥 sin 𝑥 + cos 𝑥 +𝐶 STEP 1: Decide which thing will be 𝑢 (and which 𝑑𝑣 𝑑𝑥 ). You’re about to differentiate 𝑢 and integrate 𝑑𝑣 𝑑𝑥 , so the idea is to pick them so differentiating 𝑢 makes it ‘simpler’, and 𝑑𝑣 𝑑𝑥 can be integrated easily. 𝑢 will always be the 𝑥 𝑛 term UNLESS one term is ln 𝑥 . ? STEP 2: Find 𝑑𝑢 𝑑𝑥 and 𝑣. STEP 3: Use the formula. ? I just remember it as “𝑢𝑣 minus the integral of the two new things timesed together”

Another Example ? Find 𝑥 2 ln 𝑥 𝑑𝑥 Q 𝑢= ln 𝑥 𝑑𝑣 𝑑𝑥 = 𝑥 2 𝑑𝑢 𝑑𝑥 = 1 𝑥 𝑣= 1 3 𝑥 3 𝑥 2 ln 𝑥 𝑑𝑥 = 1 3 𝑥 3 ln 𝑥 − 1 3 𝑥 2 𝑑𝑥 = 𝟏 𝟑 𝒙 𝟑 𝐥𝐧 𝒙 − 𝟏 𝟗 𝒙 𝟑 +𝑪 This time, we choose ln 𝑥 to be the 𝑢 because it differentiates nicely. STEP 1: Decide which thing will be 𝑢 (and which 𝑑𝑣 𝑑𝑥 ). STEP 2: Find 𝑑𝑢 𝑑𝑥 and 𝑣. STEP 3: Use the formula.

IBP twice!  Q Find 𝑥 2 𝑒 𝑥 𝑑𝑥 ? 𝑢= 𝑥 2 𝑑𝑣 𝑑𝑥 = 𝑒 𝑥 𝑑𝑢 𝑑𝑥 =2𝑥 𝑣= 𝑒 𝑥 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − 2𝑥 𝑒 𝑥 𝑑𝑥 We have to apply IBP again! 𝑢=2𝑥 𝑑𝑣 𝑑𝑥 = 𝑒 𝑥 𝑑𝑢 𝑑𝑥 =2 𝑣= 𝑒 𝑥 2𝑥 𝑒 𝑥 𝑑𝑥 =2𝑥 𝑒 𝑥 − 2 𝑒 𝑥 𝑑𝑥 =2𝑥 𝑒 𝑥 −2 𝑒 𝑥 Therefore 𝑥 2 𝑒 𝑥 𝑑𝑥 = 𝑥 2 𝑒 𝑥 − 2𝑥 𝑒 𝑥 −2𝑥 𝑒 𝑥 +𝐶 = 𝑥 2 𝑒 𝑥 −2𝑥 𝑒 𝑥 +2 𝑒 𝑥 +𝐶 Bro Tip: I tend to write out my working for any second integral completely separately, and then put the result back into the original integral later.

Test Your Understanding Q Find 𝑥 2 sin 𝑥 𝑑𝑥 ? =2𝑥 sin 𝑥 − 𝑥 2 cos 𝑥 +2 cos 𝑥 +𝐶

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

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

Exercise 6G ? ? ? ? ? ? ? ? ? ? ? ? ? ? 1 3 a a c c e 4 a e c 2 e a g 𝑥 2 𝑒 −𝑥 𝑑𝑥=− 𝒆 −𝒙 𝒙 𝟐 −𝟐𝒙 𝒆 −𝒙 −𝟐 𝒆 −𝒙 +𝑪 12 𝑥 2 3+2𝑥 5 𝑑𝑥 = 𝒙 𝟐 𝟑+𝟐𝒙 𝟔 − 𝟏 𝟕 𝒙 𝟑+𝟐𝒙 𝟕 + 𝟏 𝟏𝟏𝟐 𝟑+𝟐𝒙 𝟖 +𝑪 𝑥 2 2 sec 2 𝑥 tan 𝑥 𝑑𝑥 = 𝒙 𝟐 𝐬𝐞𝐜 𝟐 𝒙 −𝟐𝒙 𝐭𝐚𝐧 𝒙 +𝟐 𝐥𝐧 𝐬𝐞𝐜 𝒙 +𝑪 0 ln 2 𝑥 𝑒 2𝑥 𝑑𝑥=𝟐 𝐥𝐧 𝟐 − 𝟑 𝟒 0 𝜋 2 𝑥 cos 𝑥 𝑑𝑥 = 𝝅 𝟐 −𝟏 0 1 4𝑥 1+𝑥 3 𝑑𝑥=𝟗.𝟖 0 𝜋 3 sin 𝑥 ln⁡( sec 𝑥) 𝑑𝑥 = 𝟏 𝟐 𝟏− 𝐥𝐧 𝟐 𝑥 sin 𝑥 𝑑𝑥 =−𝒙 𝐜𝐨𝐬 𝒙 + 𝐬𝐢𝐧 𝒙 +𝑪 𝑥 sec 2 𝑥 𝑑𝑥 =𝒙 𝐭𝐚𝐧 𝒙 − 𝐥𝐧 𝐬𝐞𝐜 𝒙 +𝑪 𝑥 sin 2 𝑥 𝑑𝑥 =−𝒙 𝐜𝐨𝐭 𝒙 + 𝐥𝐧 𝐬𝐢𝐧 𝒙 +𝑪 𝑥 2 ln 𝑥 𝑑𝑥 = 𝟏 𝟑 𝒙 𝟑 𝐥𝐧 𝒙 − 𝟏 𝟗 𝒙 𝟑 +𝑪 ln 𝑥 𝑥 3 𝑑𝑥 =− 𝐥𝐧 𝒙 𝟐 𝒙 𝟐 − 𝟏 𝟒 𝒙 𝟐 +𝑪 𝑥 2 +1 ln 𝑥 𝑑𝑥 = 𝟏 𝟑 𝒙 𝟑 𝐥𝐧 𝒙 − 𝟏 𝟗 𝒙 𝟑 +𝒙 𝐥𝐧 𝒙 −𝒙 3 a ? a ? c ? ? c e ? 4 ? a Bonus Question: arcsin 𝑥 𝑑𝑥 = 𝟏− 𝒙 𝟐 +𝒙 𝐚𝐫𝐜𝐬𝐢𝐧 𝒙 +𝑪 (hint: 𝑑𝑣 𝑑𝑥 =1) e ? ? c ? ? ? 2 e a ? ? g c ? You will need the following standard results (given in your formula booklet). We’ll prove them later. ! tan 𝑥 𝑑𝑥 = ln sec 𝑥 +𝐶 sec 𝑥 𝑑𝑥 = ln sec 𝑥+ tan 𝑥 +𝐶 cot 𝑥 𝑑𝑥 = ln s𝑖𝑛 𝑥 +𝐶 cosec 𝑥 𝑑𝑥 = ln cosec 𝑥+ cot 𝑥 +𝐶 e