Outline. 1 Integration by Substitution: The Technique. 2 Integration by Substitution: Worked Examples. 3 Integration by Parts: The Technique

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1 MS2: IT Mathematics Integration Two Techniques of Integration John Carroll School of Mathematical Sciences Dublin City University Integration by Substitution: The Technique Integration by Substitution: The Technique Learning from Differentiation Integration by Substitution Learn from Differentiation To differentiate y = e 3 we substitute u = 3 and differentiate using the Chain Rule. We have y = e u u = 3 dy dy = e u = 3 2 = dy = eu (3 2) = 3 2 e 3 Integration (3/4) MS2: IT Mathematics John Carroll 3 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 4 / 37

2 Integration by Substitution: The Technique Learning from Differentiation Integration by Substitution: The Technique Learning from Differentiation Integration by Substitution Applying this to Integration y = e 3 dy = 3 2 e 3 From the above, we therefore know that 3 2 e 3 = e 3 How could we calculate the integral directly? The reason that it is difficult to calculate 3 2 e 3 is similar to that which makes it difficult to calculate the derivative of e 3. Integration (3/4) MS2: IT Mathematics John Carroll 5 / 37 Integration by Substitution Find 3 2 e 3 We try the same substitution again, so u = 3 = 3 2 = 3 2 Now replace the s under the integral (including ) by the u s (and ): 3 2 e 3 = e = e u which gives the same answer as before. = e u = e 3 Integration (3/4) MS2: IT Mathematics John Carroll 6 / 37 Integration by Substitution: The Technique Learning from Differentiation Integration by Substitution Technique: What is the best choice for u? The following are some useful tips: Let u = the most complicated epression, particularly if that epression is the argument of another function. This is the same choice of u that one would use if differentiating using the chain rule (e.g., choosing u = 3 above). Let u = f () if f () is also present under the integral. If the integrand (function to be differentiated) is a quotient, let u = the denominator. These hints may be contradictory and they are not foolproof. However, they are sufficient to tackle a large class of integrals and, with enough practice, one can develop an intuition for the correct strategy. Integration (3/4) MS2: IT Mathematics John Carroll 7 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 8 / 37

3 We let u = Replacing the s by u s: = = 5 4 = = 5 u = 5 u = 5 log u = 5 log( 5 + 2) Note: Always write the final answer in terms of the original variable (in this case ). Integration (3/4) MS2: IT Mathematics John Carroll 9 / Note that the epression which is the argument of the square root is a good candidate for substitution. Accordingly, Let u = = 4 3 = 4 3 Look to arrange 4 3 (= ) in one group in the integral: = u = 4 = u /2 4 = u +/2 4 + /2 = u3/2 = (6 + 4 ) 3/2 6 6 Integration (3/4) MS2: IT Mathematics John Carroll 0 / 37 log If f () = log, then f () = is present in the integrand. Therefore, put u = log = log = log = = u = u2 2 = (log )2 2 3 ( 4 + 4) 5 Let u = 4 + 4, the argument of the fifth power. Then = 4 3 = 4 3 Group 4 3 (= ) together and substitute: 3 ( 4 + 4) 5 = ( 4 + 4) 5. 3 = u 5 4 = u 5 = u = ( 4 + 4) 6 24 Integration (3/4) MS2: IT Mathematics John Carroll / 37 Integration (3/4) MS2: IT Mathematics John Carroll 2 / 37

4 2 + 2 = u 2 = 2 u Let u = 2 + 2, the argument of the square root. Thus = 2 = 2 Now, we replace (and ) by u (and ): = u 2 = 2 u Solution (Cont d) We now replace by epressing it in terms of u: = 2 (u 2), so = 2 (u 2) u = ( ) ( ) u 2 u = u 3/2 2u/ /2 /2 = 6 u3/2 u /2 = 6 (2 + 2)3/2 (2 + 2) /2 Integration (3/4) MS2: IT Mathematics John Carroll 3 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 4 / 37 Integration by Substitution: Trogonometry Eamples Integration by Substitution: Trogonometry Eamples sin 6 We let u = 6, the argument of the sin function: Therefore: sin 6 = = 6 = 6 sin u 6 = 6 sin u sin 2 + cos Since the integrand is a quotient, we put u = 2 + cos, so = sin. Therefore, we have sin sin 2 + cos = 2 + cos = u = log u = log(2 + cos ) = 6 ( cos u) = cos 6 6 Integration (3/4) MS2: IT Mathematics John Carroll 5 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 6 / 37

5 Integration by Substitution: Trogonometry Eamples Integration by Substitution: Trogonometry Eamples sin cos 3 Let u = cos, the argument of the cube. Then = sin = sin Group sin (= ) together and substitute: sin cos 3 = cos 3 sin = u 3 ( ) = u 3 = u4 4 = cos4 4 tan 5 cos 2 We solve as follows: d tan = sec2, = sec 2 = tan 5 cos 2 = so put u = tan to obtain cos 2 u 5 = u6 6 = tan6 6 Integration (3/4) MS2: IT Mathematics John Carroll 7 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 8 / 37 Integration by Parts: The Technique Integration by Parts: The Technique Learning from Differentiation Integration by Parts The Link to Differentiaton We have seen how integration by substitution corresponds to the chain rule for differentiation. It is reasonable to ask if there is a rule of integration corresponding to the proct rule? The answer to this question is yes and the rule is that of integration by parts. This is a rule which is useful for calculating the integrals of procts: u dv = uv v Integration (3/4) MS2: IT Mathematics John Carroll 9 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 20 / 37

6 Integration by Parts: Worked Eamples Integration by Parts: Worked Eamples Integration by Parts: u dv = uv v e 2 We will let u = and dv = e 2 so that e 2 = u dv We also need and v: u = i.e. we differentiate u to get. Also dv = e 2 v = = = e 2 = 2 e2 i.e. we integrate dv to get v. Integration (3/4) MS2: IT Mathematics John Carroll 2 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 22 / 37 Integration by Parts: Worked Eamples Integration by Parts: Worked Eamples Integration by Parts: u dv = uv v Let so that e 2 (Cont d) u = and dv = e 2 = and v = e 2 = 2 e2 Now, substitute u, v, and dv into the integration by parts formula to obtain: e 2 = u dv = uv v = 2 e2 2 e2 = 2 e2 4 e2 Integration by Parts: u dv = uv v cos Choose u = and dv = cos, so that Inspecting the right-hand side of the integration by parts formula, we also need and v: u = = dv = cos v = = cos = sin Substitute u, v, and dv into the formula to obtain: cos = u dv = uv v = sin sin = sin + cos Integration (3/4) MS2: IT Mathematics John Carroll 23 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 24 / 37

7 Integration by Parts: Worked Eamples Integration by Parts: Worked Eamples Integration by Parts: u dv = uv v Integration by Parts: u dv = uv v 2 log Important Question How do we know in practice what to choose for u and dv? To answer the question, we first consider another eample where we make our choice by trial and error. We then present a set of guidelines to assist in the decision-making process. We could choose either u = 2 or u = log Which should one try? (a) u = 2 : Then dv = log, so v = log. This is not listed in the formulae and tables. We will therefore abandon this approach. Integration (3/4) MS2: IT Mathematics John Carroll 25 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 26 / 37 Integration by Parts: Worked Eamples Integration by Parts: Worked Eamples LATE Guidelines Integration by Parts: u dv = uv v Integration by Parts: u dv = uv v 2 log (Cont d) (b) u = log : Then dv = 2, so v = 2 = 3 3 and = = Answer to the Important Question To decide how u should be chosen in general, we advocate that the choice should be made in the following order: Substitute these epressions into the integration by parts formula: 2 log = u dv = uv v = log = 3 3 log 2 = log 9 3 L Logarithm A Algebraic: powers of, e.g., 3 T Trigonometric E Eponential Integration (3/4) MS2: IT Mathematics John Carroll 27 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 28 / 37

8 Integration by Parts: Worked Eamples LATE Guidelines Integration by Parts: Worked Eamples Using the LATE Guidelines Integration by Parts: u dv = uv v LATE Guidelines The guideline has been used in all the eamples: e 2 : A, e 2 E LATE u = (A), so dv = e cos : A, cos T LATE u = (A), so dv = cos. 2 log 2 A, log L LATE u = log (L), so dv = 2. This rule gives very reliable guidance, but occasionally it does not work. Sometimes, we must integrate by parts successively to get an answer as the following eample will illustrate. Integration by Parts: u dv = uv v 2 cos According to our notation, this is of type LATE u = 2 (A), so dv = cos (T). Therefore, we differentiate u and integrate dv to obtain: = 2, v = cos = sin Substitute u, v,, dv into the formula: 2 cos = u dv = uv v = 2 sin sin 2 = 2 sin 2 We see that we cannot calculate the last integral directly. sin Integration (3/4) MS2: IT Mathematics John Carroll 29 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 30 / 37 Integration by Parts: Worked Eamples Using the LATE Guidelines Integration by Parts: Worked Eamples Using the LATE Guidelines Integration by Parts: u dv = uv v 2 cos = 2 sin 2 sin We must use the integration by parts formula again for sin which is again of type LATE u = (A), so dv = sin (T). We differentiate u and integrate dv to obtain: =, v = sin = cos Substituting into the formula yields: sin = cos ( cos ) = cos + sin Integration (3/4) MS2: IT Mathematics John Carroll 3 / 37 Integration by Parts: u dv = uv v 2 cos (Cont d) 2 cos = 2 sin 2 sin sin = cos + sin Substituting the second result into the first yields: 2 cos = 2 sin 2 sin = 2 sin 2( cos + sin ) = 2 sin + 2 cos 2 sin Integration (3/4) MS2: IT Mathematics John Carroll 32 / 37

9 Two Definite Integral Eamples Two Definite Integral Eamples Substitution Eample Definite Integrals: Integration by Substitution Eample 4 Substitution u = = 6 = 6 = 6 This means that = 6 u Integration (3/4) MS2: IT Mathematics John Carroll 33 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 34 / 37 Two Definite Integral Eamples Substitution Eample Two Definite Integral Eamples Integration-by-parts Eample Definite Integrals: Integration by Substitution Eample Substitution u = leads to = 6 u Change the limits of integration = u =3 2 + = 4 = 4 u =3 2 + = 49 Therefore = = 6 = ( ) = u 49 [ln u]49 4 = (ln 49 ln 4) 6 4 u ( ) 6 Definite Integrals: Integration by Parts Eample 3 r 3 ln r dr Integration by Parts u = ln r = r dr dv = r 3 dr v = 4 r 4 Integration (3/4) MS2: IT Mathematics John Carroll 35 / 37 Integration (3/4) MS2: IT Mathematics John Carroll 36 / 37

10 Two Definite Integral Eamples Integration-by-parts Eample Definite Integrals: Integration by Parts Eample 3 r 3 ln r dr with u = ln r and dv = r 3 dr v = 4 r 4 Using the integration by parts formula 3 u dv = r 3 ln r dr = = [ 4 r 4 ln r ] 3 3 = 8 4 ln [ uv 4 r 3 dr [ 4 r 4 ] 3 ] 3 v r= = 8 4 ln 3 8 (8 ) = 6 4 ln 3 5 Integration (3/4) MS2: IT Mathematics John Carroll 37 / 37