MTH240 – CALCULUS II LECTURE 1 Substitution and Integration By Parts Volume 2 Chapter 3, Section 3.1 Winter 2024 Copyright © (2024) by F. K. Duah Learning Outcomes After studying the material covered in this session, you should be able to : • Explain the concept of integration by parts. • Recognize when to use integration by parts. • Use integration-by-parts formula to solve integration problems. Copyright © (2024) by F. K. Duah 2 Pre-requisites from Calculus I It is reasonable to expect the reader to recall the following integrals and their results. π 1) β«= π₯π π₯ Χ¬β¬ π₯ π+1 π+1 + π, π ≠ − 1 2) β«π )π₯(π π Χ¬β¬′ π₯ ππ₯ = πΉ π(π₯) + π where πΉ(β) is the antiderivative of π ⋅ . 3) β«π₯ π Χ¬β¬ 4) β«π π Χ¬β¬ 5) β«π Χ¬β¬ 6) π₯ π π₯ π′ π₯ β«π₯πΧ¬β¬ π ′ π π₯ ππ₯ = π ′ π₯ ππ₯ = π π ′ π π₯ ππ₯ = π π₯ π+1 π+1 π₯ ππ π₯ ln π + π, π ≠ −1 +π + π, π > 0, π ≠ 1 ππ₯ = ln π π₯ + π Copyright © (2024) by F. K. Duah 3 Pre-requisites from Calculus I The following integrals can all easily be evaluated by inspection or by u-substitution even though they involve the products of two functions in the integrand. ΰΆ± 18π₯ 2 4 6π₯ 3 + 13 ππ₯ ΰΆ± 8π₯ − 1 π 4π₯ 2 −π₯ ππ₯ Copyright © (2024) by F. K. Duah 2π₯ 3 + 1 ΰΆ± 4 ππ₯ π₯ + 2π₯ 3 ΰΆ± 1− 1 cos π₯ − ln π₯ ππ₯ π₯ 4 Integration by Parts When solving an engineering problem that require evaluation of an integral, we may find products of functions that are difficult to integrate unless they are transformed to integrals that are easier to integrate than the original integrals. Examples of such integrals are: ΰΆ± π₯π π₯ ππ₯ ΰΆ± π₯ 2 π π₯ ππ₯ ΰΆ± π₯ sin π₯ ππ₯ The technique that transforms such integrals into easier integrals is what is referred to as Integration by Parts. In the next few slides, we look at the theory behind the technique. Copyright © (2024) by F. K. Duah 5 Theory of Integration by Parts Integration by parts is based on the Product Rule of Differentiation. Suppose π’ and π£ are differentiable functions. The product rule for differentiation is given by: π ππ’ ππ£ π’π£ = π£ +π’ . ππ₯ ππ₯ ππ₯ If we take the antiderivative of both sides of the above equation, we obtain π ππ’ ππ£ ΰΆ± π’π£ ππ₯ = ΰΆ± π£ ππ₯ + ΰΆ± π’ ππ₯ . ππ₯ ππ₯ ππ₯ Copyright © (2024) by F. K. Duah 6 Theory of Integration by Parts We know from the previous slide that π ππ’ ππ£ ΰΆ± π’π£ ππ₯ = ΰΆ± π£ ππ₯ + ΰΆ± π’ ππ₯. ππ₯ ππ₯ ππ₯ The antiderivative of the left hand is given by ππ’ ππ£ π’π£ = ΰΆ± π£ ππ₯ + ΰΆ± π’ ππ₯. ππ₯ ππ₯ When we re-arrange the above equation we obtain, ππ£ ππ’ ΰΆ±π’ ππ₯ = π’π£ − ΰΆ± π£ ππ₯ . ππ₯ ππ₯ The above equation is the integration by parts formula. Copyright © (2024) by F. K. Duah 7 Integration by Parts Recall that the integration by parts formula is ππ£ ππ’ ΰΆ±π’ ππ₯ = π’π£ − ΰΆ± π£ ππ₯. ππ₯ ππ₯ For ease of application, the integration by parts formula is typically rewritten in terms of differentials as ΰΆ± π’ ππ£ = π’π£ − ΰΆ± π£ ππ’ . Copyright © (2024) by F. K. Duah 8 ΰΆ± π’ ππ£ = π’π£ − ΰΆ± π£ ππ’ Activity 1 Integration By Parts Evaluate β« ππ₯ Χ¬β¬−5π₯ ππ₯. Solution Let π’ = π₯ and ππ£ = π −5π₯ ππ₯ ππ’ = ππ₯ and π£ = β«Χ¬β¬ π −5π₯ ππ₯ ΰΆ± π₯π −5π₯ ππ₯ = 1 −5π₯ − π 5 1 −5π₯ 1 −5π₯ =π₯ − π −ΰΆ± − π ππ₯ 5 5 π₯ −5π₯ 1 5π₯ =− π − π +π 5 25 Copyright © (2024) by F. K. Duah 9 Integration by Parts Choosing u and dv. Recall the integration by parts formula ΰΆ± π’ ππ£ = π’π£ − ΰΆ± π£ ππ’ There is an order of precedence for the choices of π’ and ππ£ and this goes by the acronym LIATE. L – Logarithm I – Inverse Trigonometric Functions A – Algebraic Functions T – Trigonometric Functions E – Exponential Functions Thus, we choose the function which appears first in the list LIATE and let it be π’. We let ππ£ be equal to the other function and ππ₯. This is only for guidance and my not always work or without some preliminary work. Copyright © (2024) by F. K. Duah 10 ΰΆ± π’ ππ£ = π’π£ − ΰΆ± π£ ππ’ Activity 2 Evaluate β« π₯ Χ¬β¬sin π₯ ππ₯ . Solution Let π’ = π₯ and ππ£ = sin π₯ ππ₯ ππ’ = ππ₯ and π£ = β« Χ¬β¬sin π₯ ππ₯ = − cos π₯ ΰΆ± π₯ sin π₯ ππ₯ = π₯ − cos π₯ − ΰΆ± −cos π₯ ππ₯ = −π₯ cos π₯ + sin π₯ + π Copyright © (2024) by F. K. Duah 11 ΰΆ± π’ ππ£ = π’π£ − ΰΆ± π£ ππ’ Activity 3 Evaluate β« Χ¬β¬5π₯ + 1 cos 2π₯ ππ₯. Solution Let π’ = 5π₯ + 1 and ππ£ = cos 2π₯ ππ₯ 1 ππ’ = 5ππ₯ and π£ = β« Χ¬β¬cos 2π₯ ππ₯ = sin 2π₯ 2 ΰΆ± 5π₯ + 1 cos 2π₯ ππ₯ = 5π₯ + 1 1 1 sin 2π₯ − ΰΆ± sin 2π₯ 2 2 5ππ₯ 1 5 = 5π₯ + 1 sin 2π₯ + cos 2π₯ 2 4 Copyright © (2024) by F. K. Duah 12 ΰΆ± π’ ππ£ = π’π£ − ΰΆ± π£ ππ’ Activity 4 Evaluate β« Χ¬β¬ln π₯ ππ₯. Solution Let π’ = ln π₯ and ππ£ = 1 ππ₯ 1 ππ’ = ππ₯ and π£ = β«π₯ = π₯π Χ¬β¬ π₯ 1 ΰΆ± ln π₯ ππ₯ = ln π₯ π₯ − ΰΆ± π₯ ππ₯ π₯ = π₯ ln π₯ − π₯ + π. Copyright © (2024) by F. K. Duah 13 ΰΆ± π’ ππ£ = π’π£ − ΰΆ± π£ ππ’ Activity 5 Evaluate β« π₯ Χ¬β¬2 π π₯ ππ₯. Solution Let π’ = π₯ 2 and ππ£ = π π₯ , ππ’ = 2π₯ππ₯ and π£ = β«π₯ π = π₯π π₯ π Χ¬β¬ ΰΆ± π₯ 2 π π₯ ππ₯ = π₯ 2 π π₯ − ΰΆ± π π₯ 2π₯ ππ₯ = π₯ 2 π π₯ − 2 ΰΆ± π₯π π₯ ππ₯ Integrate β« π₯π π₯ ππ₯ Χ¬β¬using integration by parts. Let π’ = π₯ and ππ£ = π π₯ , ππ’ = ππ₯ and π£ = β«π₯ π = π₯π π₯ π Χ¬β¬ ΰΆ± π₯π π₯ ππ₯ = π₯π π₯ − ΰΆ± π π₯ ππ₯ = π₯π π₯ − π π₯ ΰΆ± π₯ 2 π π₯ ππ₯ = π₯ 2 π π₯ − ΰΆ± π π₯ 2π₯ ππ₯ = π₯ 2 π π₯ − 2 π₯π π₯ − π π₯ + π = π₯ 2 π π₯ − 2π₯π π₯ + 2π π₯ + π. This technique will work for any integral of the form β« π₯π π₯ π π π₯ Χ¬β¬, were π is a positive integer. Copyright © (2024) by F. K. Duah 14 ΰΆ± π’ ππ£ = π’π£ − ΰΆ± π£ ππ’ Activity 6 Evaluate β« π₯ π Χ¬β¬cos π₯ ππ₯. Solution Let π’ = cos π₯ and ππ£ = π π₯ , ππ’ = −sin π₯ ππ₯ and π£ = β«π₯ π = π₯π π₯ π Χ¬β¬ ΰΆ± π π₯ cos π₯ ππ₯ = cos π₯ π π₯ − ΰΆ± π π₯ −sin π₯ ππ₯ = π π₯ cos π₯ − ΰΆ± π π₯ −sin π₯ ππ₯ = π π₯ cos π₯ + ΰΆ± π π₯ sin π₯ ππ₯ Integrate β« π₯ π Χ¬β¬sin π₯ ππ₯ using integration by parts. Let π’ = sin π₯ and ππ£ = π π₯ , ππ’ = cos π₯ ππ₯ and π£ = β«π₯ π = π₯π π₯ π Χ¬β¬ ΰΆ± π π₯ sin π₯ ππ₯ = sin π₯ π π₯ − ΰΆ± π π₯ cos π₯ ππ₯ = π π₯ sin π₯ − ΰΆ± π π₯ cos π₯ ππ₯ ΰΆ± π π₯ cos π₯ ππ₯ = π π₯ cos π₯ + π π₯ sin π₯ − ΰΆ± π π₯ cos π₯ ππ₯ π π₯ cos π₯ + π π₯ sin π₯ ΰΆ± π cos π₯ ππ₯ = +π 2 π₯ Copyright © (2024) by F. K. Duah 15 Further Review of Calculus I Activities 7 to 20 are problems aimed at helping you consolidate your knowledge of the techniques of integration learned in MTH140 of Calculus I. The activities require knowledge of : • The Power Rule • Substitution • Chain Rule You are advised to ensure that you are proficient in evaluating integrals of the kind presented in Activities 7 to 20. But before we look at these, a quick recap of some calculus facts. Copyright © (2024) by F. K. Duah 16 Review of Calculus I - Derivatives To progress smoothly through Calculus II, you should be able to recall and evaluate derivatives of the following forms: π π π=π π = π′ π π π π¦ = π, where π is constant. ππ¦ =0 π¦ = π, where π is constant ππ₯ ππ¦ π¦ = π₯π = ππ₯ π−1 ππ₯ ππ¦ π¦ = ππ₯ = ππ₯ ππ₯ ππ¦ 1 π¦ = ln π₯ = ππ₯ π₯ ππ¦ π¦ = sin π₯ = cos π₯ ππ₯ ππ¦ π¦ = cos π₯ = − sin π₯ ππ₯ Copyright © (2024) by F. K. Duah 17 Review of Calculus I - Derivatives To progress smoothly through Calculus II, you should be able to recall and evaluate derivatives of the following forms: π=π π π¦ = tan π₯ π¦ = cot π₯ π¦ = sec π₯ π¦ = csc π₯ Copyright © (2024) by F. K. Duah π π = π′ π π π ππ¦ = sec 2 π₯ ππ₯ ππ¦ = −csc 2 π₯ ππ₯ ππ¦ = sec π₯ tan π₯ ππ₯ ππ¦ = − csc π₯ cot π₯ ππ₯ 18 Review of Calculus I - Derivatives To progress smoothly through Calculus II, you should be able to recall and evaluate derivatives of the following forms: π=π π π¦ = sin−1 π₯ π¦ = cos−1 π₯ π¦ = tan−1 π₯ Copyright © (2024) by F. K. Duah π π = π′ π π π 1 ππ¦ = ππ₯ 1 − π₯2 ππ¦ 1 =− ππ₯ 1 − π₯2 ππ¦ 1 = ππ₯ 1 + π₯ 2 19 Review of Calculus I - Integrals To progress smoothly through Calculus II, you should be able to recall and evaluate integrals of the following forms: 1) β« π₯ = π₯π Χ¬β¬+ π 2) β«Χ¬β¬ π₯ π ππ₯ 3) β«π₯ π Χ¬β¬ = π₯ π+1 π+1 π ′ π + π, π ≠ − 1 π₯ ππ₯ = 4) β«π π Χ¬β¬ π₯ π ′ π₯ ππ₯ = π π 5) β«π π Χ¬β¬ π₯ π ′ π₯ ππ₯ = 6) π′ π₯ β«π₯πΧ¬β¬ π π₯ π+1 π+1 π₯ ππ π₯ ln π + π, π ≠ −1 +π + π, π > 0, π ≠ 1 ππ₯ = ln π π₯ + π Copyright © (2024) by F. K. Duah 20 Review of Calculus I - Integrals To be progress smoothly through Calculus II, you should be able to recall and apply the follow integrals. Some examples are given where appropriate. 7) β« Χ¬β¬sin π π₯ π ′ π₯ ππ₯ = −cos π π₯ 8)β« Χ¬β¬cos π π₯ π ′ π₯ ππ₯ = sin π π₯ +π +π 9) β« Χ¬β¬sec 2 π π₯ π ′ π₯ ππ₯ = tan π π₯ +π 10) β« Χ¬β¬csc 2 π π₯ π ′ π₯ ππ₯ = −cot π π₯ +π 11)β« Χ¬β¬sec π π₯ tan π π₯ π ′ π₯ ππ₯ = sec π π₯ 12)β« Χ¬β¬csc π π₯ cot π π₯ π ′ π₯ ππ₯ = −csc π π₯ 13)β« Χ¬β¬tan π π₯ π ′ π₯ ππ₯ = ln sec π π₯ Copyright © (2024) by F. K. Duah +π +π +π 21 Review of Calculus I - Integrals To be progress smoothly through Calculus II, you should be able to recall and apply the follow integrals. Some examples are given where appropriate. 1 1 14) β«π Χ¬β¬2+π₯2 ππ₯ = π tan−1 15) β«Χ¬β¬ 16) β«Χ¬β¬ 1 π2 −π₯ ππ₯ = sin−1 2 1 π₯ π2 −π₯ 1 π₯ π π₯ π ππ₯ = π sec −1 2 Copyright © (2024) by F. K. Duah + π, π > 0 + π, −π ≤ π₯ ≤ π π₯ π + π, π > 0 22 π+1 ΰΆ± π π₯ Activity 7 π π₯ π π₯ ππ₯ = π+1 π ′ + π , π ≠ −1 Evaluate β« Χ¬β¬2π₯ π₯ 2 + 1 5 ππ₯. Solution ΰΆ± 2π₯ π₯ 2 + 1 5 ππ₯ = ΰΆ± π₯ 2 + 1 π₯2 + 1 = 6 Copyright © (2024) by F. K. Duah 5 2π₯ ππ₯ 6 + π. 23 π+1 ΰΆ± π π₯ Activity 8 π π₯ π π₯ ππ₯ = π+1 π ′ + π , π ≠ −1 Evaluate β« π₯ Χ¬β¬3π₯ 2 + 1 5 ππ₯. Solution ΰΆ± 2π₯ π₯ 2 + 1 5 ππ₯ = ΰΆ± π₯ 2 + 1 π₯2 + 1 = 6 Copyright © (2024) by F. K. Duah 5 2π₯ ππ₯ 6 + π. 24 π+1 π π₯ π ′ ΰΆ± π π₯ π π₯ ππ₯ = π+1 Activity 9 + π , π ≠ −1 Evaluate β« Χ¬β¬sin3 π₯ cos π₯ ππ₯. Solution β«Χ¬β¬ sin3 π₯ cos ππ₯ = β« Χ¬β¬sin π₯ Copyright © (2024) by F. K. Duah 3 cos π₯ ππ₯ = sin4π₯ 4 + π. 25 π(π₯) π ΰΆ± π π(π₯) π ′ π₯ ππ₯ = +π ln π Activity 10 Evaluateβ«Χ¬β¬ π₯3 4 π₯ 5 ππ₯ . Solution ΰΆ± π₯3 4 π₯ 5 1 π₯4 ππ₯ = ΰΆ± 5 4 = Copyright © (2024) by F. K. Duah 4 π₯ 15 4 ln5 4π₯ 3 ππ₯ + π. 26 π+1 Activity 11 ΰΆ± π π₯ π π₯ π π₯ ππ₯ = π+1 π ′ + π , π ≠ −1 Evaluate β« Χ¬β¬cos3 π₯ sin π₯ ππ₯. Solution ΰΆ± cos3 π₯ sin π₯ ππ₯ = ΰΆ± − cos π₯ 3 − sin π₯ ππ₯ 1 = − cos4 π₯ + π. 4 Copyright © (2024) by F. K. Duah 27 π′ π₯ ΰΆ± ππ₯ = ln π π₯ + π π π₯ Activity 12 Evaluate β« Χ¬β¬cot π₯ ππ₯. Solution cos π₯ ΰΆ± cot π₯ ππ₯ = ΰΆ± ππ₯ sin π₯ = ln sin π₯ + π Copyright © (2024) by F. K. Duah 28 ΰΆ± sin π π₯ π ′ π₯ ππ₯ = −cos π π₯ +π Activity 13 Evaluateβ« Χ¬β¬sin sec π₯ sec π₯ tan π₯ ππ₯. Solution ΰΆ± sin sec π₯ sec π₯ tan π₯ ππ₯ = −cos sec π₯ + π. Copyright © (2024) by F. K. Duah 29 ΰΆ± cos π π₯ π ′ π₯ ππ₯ = sin π π₯ +π Activity 14 Evaluate β« Χ¬β¬cos 3π₯ 3π₯ ln 3 ππ₯. Solution ΰΆ± cos 3π₯ 3π₯ ln 3 ππ₯ = sin 3π₯ + π. Copyright © (2024) by F. K. Duah 30 ΰΆ± sec 2 π π₯ π ′ π₯ ππ₯ = tan π π₯ +π Activity 15 Evaluateβ« Χ¬β¬sec2 π π₯ π π₯ ππ₯. Solution ΰΆ± sec2 π π₯ π π₯ ππ₯ = tan π π₯ + π . Copyright © (2024) by F. K. Duah 31 ΰΆ± csc 2 π π₯ π ′ π₯ ππ₯ = −cot π π₯ Activity 16 +π Evaluate β« π₯ Χ¬β¬2 csc2 π₯ 3 − 5 ππ₯. Solution ΰΆ± π₯ 2 csc2 π₯3 1 − 5 ππ₯ = ΰΆ± csc2 π₯ 3 − 5 3π₯ 2 ππ₯ 3 1 = − cot π₯ 3 − 5 + π. 3 Copyright © (2024) by F. K. Duah 32 ΰΆ± sec π π₯ tan π π₯ π ′ π₯ ππ₯ = sec π π₯ Activity 17 Evaluateβ« Χ¬β¬sec lnπ₯ tan lnπ₯ +π 1 ππ₯. π₯ Solution 1 ΰΆ± sec lnπ₯ tan lnπ₯ ππ₯ = sec lnπ₯ + π. π₯ Copyright © (2024) by F. K. Duah 33 ΰΆ± csc π π₯ cot π π₯ π ′ π₯ ππ₯ = −csc π π₯ Activity 18 Evaluateβ« Χ¬β¬csc π₯ cot π₯ 1 2 π₯ +π ππ₯ . Solution ΰΆ± csc Copyright © (2024) by F. K. Duah π₯ cot π₯ 1 2 π₯ ππ₯ = −csc π₯ + π. 34 ΰΆ± tan π π₯ π ′ π₯ ππ₯ = ln sec π π₯ Activity 19 +π Evaluateβ« π₯ Χ¬β¬5 tan π₯ 6 ππ₯ . Solution ΰΆ± π₯ 5 tan π₯6 1 ππ₯ = ΰΆ± tan π₯ 6 6π₯ 5 ππ₯ 6 1 = ln sec π₯ 6 + π. 6 Copyright © (2024) by F. K. Duah 35 ΰΆ± tan π π₯ π ′ π₯ ππ₯ = ln sec π π₯ Activity 20 +π Evaluateβ« Χ¬β¬8π₯ 2 tan π₯ 3 ππ₯. Solution ΰΆ± 8π₯ 2 tan π₯3 8 ππ₯ = ΰΆ± tan π₯ 3 3π₯ 2 ππ₯ 3 8 = ln sec π₯ 3 + π. 3 Copyright © (2024) by F. K. Duah 36 Recap In this session, we have • introduced and discussed the theory of integration by parts, and looked at some examples of integration by parts, • reviewed some common integrals, and • reviewed how to integrate some functions mentally by inspection. Next lecture we will develop our understanding of integration by parts and the application of the reduction formula to solve some problems. Copyright © (2024) by F. K. Duah 37 The End Thank You For Coming Copyright © (2024) by F. K. Duah 38
