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14CHAPTER 1. TECHNIQUES OF INTEGRATIONSolution 1.16. Let θ sin 1 x. So, x sin θ and d x cos θ d θ. Thus,ZZ¡ 1 2sin x d x θ 2 cos θ d θ (let u θ 2 and d v cos θ d θ)Z2 θ sin θ 2θ sin θ d θ (let u 2θ and d v sin θ d θ)Z2 θ sin θ 2θ cos θ 2 cos θ d θ θ 2 sin θ 2θ cos θ 2 sin θ Cp¡ 2 x sin 1 x 2 1 x 2 sin 1 x 2x C1θZExample 1.17. Evaluatep1 x2x 2e xd x.(x 2)2Solution 1.17. let u x 2 e x and d v Zx1dx(x 2)2Z xx 2e xx 2e x(x 2)xe dx dx x 2(x 2)2x 2Zx 2e x xe x d xx 2x 2e x xe x e x Cx 2Exercise 1.2. Evaluate each of the following integrals.Z¡ 2 1.x x 1 e x d x. Hint: by parts, let u x 2 x 1Zpx x 5 d x. Hint: by substitution, let y x 5Zπ/82.3.0x sec2 x d x. Hint: by parts, let u x

1.3. TRIGONOMETRIC INTEGRALS15Zcos (ln x) d x. Hint: Start by substituting y ln x4.Z5.xe x sin x d x. Hint: by parts, let u xZ³ln x 6.px2 1 ³d x. Hint: by parts, let u ln x Z1 sin xxd x. Hint: Multiply by1 sin x1 sin xZln x 17.8.(ln x)2Z9.px2 1 d x.x (1 ln x)2 d x. Hint: by parts, let u (1 ln x)2Z(ln 2x) (ln x) d x. Hint: ln(ab) ln a ln b10.Z¶¡ x 1lnd x. Hint: ln ba ln a ln bx 1Zp11.12.µx tan 1px d x.1.3 Trigonometric IntegralsIn this section you will study techniques for evaluating integrals of the formZZmnsin x cos x d xandsecm tann d xwhere either m or n is a positive integer. To find anti-derivatives for theseforms, try to break them into combinations of trigonometricintegrals to whichRyou can apply the Power Rule. To break up sinm x cosn x d x into forms towhich you can apply the Power Rule, use the following identities.sin2 θ cos2 θ 11 cos(2θ)21 cos(2θ)cos2 θ 2sin2 θ

16CHAPTER 1. TECHNIQUES OF INTEGRATIONAlgorithm 1.1. Guidelines for Evaluating Integrals Involving Powers of Sineand Cosine1. If the power of the sine is odd and positive, save one sine factor andconvert the remaining factors to cosines. Then, expand and integrate.Convert to cosZOddSave for d uZ z } {z } {z } {¡ ksin2k 1 x cosn x d x sin2 xcosn x sin x d xZ¡ k 1 cos2 x cosn x sin x d x2. If the power of the cosine is odd and positive, save one cosine factor andconvert the remaining factors to sines. Then, expand and integrate.Convert to sinZOddfor d uz } { SaveZz } {¡ k z } {22k 1mmcos x d xx d x sin x cos xsin x cosZ¡ k sinm x 1 sin2 x cos x d x3. If the powers of both the sine and cosine are even and non-negative,make repeated use of the identitiessin2 x 1 cos(2x)2andcos2 x 1 cos(2x)2to convert the integrand to odd powers of the cosine. Then proceed asin guideline 2.ZExample 1.18. Evaluatesin3 x cos4 x d x.Solution 1.18. Because you expect to use the Power Rule with u cos x, save

1.3. TRIGONOMETRIC INTEGRALS17one sine factor to form d u and convert the remaining sine factors to cosines.Z34Zsin2 x cos4 x sin x d xZ¡ 1 cos2 x cos4 x sin x d xZ¡ cos4 x cos6 x sin x d xsin x cos x d x Z Let u cos x¡ 6 u u4 d u11 u 7 u 5 C7511 cos7 x cos5 x C75In the next example the power of the cosine is 3, but the power of the sineis 12 .ZExample 1.19. Findπ/2π/6cos3 xd x.psin xSolution 1.19. Because you expect to use the Power Rule with u sin x, saveone cosine factor to form d u and convert the remaining cosine factors tosines.Zπ/2π/6π/2cos2 x cos xdxpπ/6sin x Z π/2 ¡1 sin2 x cos x dxpπ/6sin xZ 1 1 ¡ u 2 1 u2 d ucos3 xdx psin xZ1/21 ³Z u1/2ZExample 1.20. Evaluate 21 u32cos4 x d x. Let u sin xp· 12 5 132 19 2 d u 2u 2 u 25201/2

18CHAPTER 1. TECHNIQUES OF INTEGRATIONSolution 1.20. Because m and n are both even and non-negative (m 0) youhi2can replace cos4 x by 1 cos(2x).2Z 1 cos(2x) 2dxcos x d x 2 Z ·1 cos(2x) cos2 (2x) dx424 ·Z1 cos(2x) 1 cos(4x) dx 428ZZZ311 dx cos(2x) d x cos(4x) d x828311 x sin(2x) sin(4x) C8432Z ·4Theorem 1.3.1. WALLIS’S FORMULAS1. If n is odd (n 3), thenZπ/20µ ¶µ ¶µ ¶ µ¶2 4 6n 1cos x d x ···.3 5 7nn2. If n is even (n 2), thenZ0π/2cosn x d x µ ¶µ ¶µ ¶ µ¶1 3 5n 1 ³π ···.2 4 6n2These formulas are also valid if cosn x is replaced by sinn x.Z π/2¡ Example 1.21. Evaluate8 cos4 x 3 sin5 x d x.0Solution 1.21. By using Wallis’s Formulas, we haveπ/2 ¡Z045 Zπ/2Zπ/2cos x d x 3sin5 x d xµ0 ¶ µ ¶ ³ µ ¶ 0µ ¶1 3 π2 4 3 82 4 23 515π 16 108 cos x 3 sin x d x 84

1.3. TRIGONOMETRIC INTEGRALS19RThe following guidelines can help you evaluate integrals of the form secm x tann x d x.Algorithm 1.2. Guidelines for Evaluating Integrals Involving Powers of Secant and Tangant1. If the power of the secant is even and positive, save a secant-squaredfactor and convert the remaining factors to tangents. Then expand andintegrate.Convert to tanEvenZZz} {2knsecx tan x d x Z Save for d u} {zz } {¡ 2 k 1sec xtann x sec2 x d x¡ k 11 tan2 xtann x sec2 x d x2. If the power of the tangent is odd and positive, save a secant-tangentfactor and convert the remaining factors to secants. Then expand andintegrate.Convert to secZOddfor d uz } { z Save} Zz } {{¡ ksec x tan x d xsecm x tan2k 1 x d x secm 1 x tan2 xZ¡ k secm 1 x sec2 x 1 sec x tan x d x3. If there are no secant factors and the power of the tangent is even andpositive, convert a tangent-squared factor to a secant-squared factor,then expand and repeat if necessary.Convert to secZZntan x d x tanZ n 2z¡ } { x tan2 x d x¡ tann 2 x sec2 x 1 d xR4. If the integral is of the form secm x d x where m is odd and positive,use integration by parts, as illustrated in Example 1.15 in the precedingsection.5. If none of the above applies, try converting to sines and cosines.

20CHAPTER 1. TECHNIQUES OF INTEGRATIONtan3 xd x.psec xZExample 1.22. EvaluateSolution 1.22. Because you expect to use the Power Rule with u sec x, savea factor of sec x tan x to form d u and convert the remaining tangent factors tosecants.Ztan3 xdx psec xZ(sec x) 2 tan3 x d xZ(sec x) 2 tan2 x sec x tan x d x 13Z 3 ¡(sec x) 2 sec2 x 1 sec x tan x d x Let u sec xZZ ³ ¡ 2 13 32 uu 1 du u 2 u 2 d u 12 3 u 2 2u 2 C3312 sec 2 x 2 sec 2 x C3Zsec4 (3x) tan3 (3x) d x.Example 1.23. FindSolution 1.23. Let u tan(3x) then d u 3 sec2 (3x)d x and you can writeZ4Z3sec (3x) tan (3x) d x sec2 (3x) tan3 (3x) sec2 (3x) d xZ¡ 1 tan2 (3x) tan3 (3x) sec2 (3x) d xZZ 3 1 ¡1 ¡ 32 1 u u du u u5 d u331 4 1 6 u u C121811 tan4 (3x) tan6 (3x) C1218 π/4ZExample 1.24. Evaluate0tan4 x d x.

1.3. TRIGONOMETRIC INTEGRALS21Solution 1.24. Because there are no secant factors, you can begin by converting a tangent-squared factor to a secant-squared factor.ZZZ¡ 422tan x d x tan x tan x d x tan2 x sec2 x 1 d xZZ22 tan x sec x d x tan2 x d xZZ¡ 2 22 tan x sec x d x sec x 1 d x 1tan3 x tan x x C3You can evaluate the definite integral as follows.π/4Z0·1tan3 x tan x xtan x d x 34 π/4 0π 2 4 3For integrals involving powers of cotangents and cosecants, you can follow a strategy similar to that used for powers of tangents and secants. Also,when integrating trigonometric functions, remember that it sometimes helpsto convert the entire integrand to powers of sines and cosines.Zsec xExample 1.25. Findd x.tan2 xSolution 1.25. Because the guidelines do not apply, try converting the integrand to sines and cosines. In this case, you are able to integrate the resultingpowers of sine and cosine as follows.Zsec xdx tan2 xZ µ1cos xZcos x¶µcos2 xsin2 x¶dxd x Let u sin x d u cos xd x2sinxZ11 du Cu2u1 C csc x Csin x Integrals involving the products of sines and cosines of two different angles occur in many applications. In such instances you can use the followingproduct-to-sum identities.

22CHAPTER 1. TECHNIQUES OF INTEGRATION1{cos [(m n)x] cos [(m n)x]}21sin(mx) cos(nx) {sin [(m n)x] sin [(m n)x]}21cos(mx) cos(nx) {cos [(m n)x] cos [(m n)x]}2ZExample 1.26. Find sin(5x) cos(4x) d x.sin(mx) sin(nx) Solution 1.26. Considering the second product-to-sum identity above, youcan writeZZ1sin(5x) cos(4x) d x (sin x sin(9x)) d x211cos(9x) C cos x 218Exercise 1.3. Evaluate the following integrals.Z1.sin5 x d xZsin5 x cos x d xZsin x tan2 x d xZx sin2 x d xZ¡ 4 tan x sec4 x d xZcos(2x)dxcos xZπ/22.3.4.5.6.7.sin12 x d x0Z8.sin( 4x) sin(3x) d x

1.4. TRIGONOMETRIC SUBSTITUTIONS231.4 Trigonometric SubstitutionsNow that you can evaluate integrals involving powers of trigonometric functions, you cansubstitutionto evaluate integrals involvingppp use trigonometricthe radicals a 2 x 2 , a 2 x 2 and x 2 a 2 . The objective with trigonometric substitution is to eliminate the radical in the integrand. You do this byusing the Pythagorean identitiescos2 θ 1 sin2 θ,sec2 θ 1 tan2 θ,tan2 θ sec2 θ 1Note 1.1. TRIGONOMETRIC SUBSTITUTION1. For integrals involvingppa 2 x 2 , let x a sin θ. Thena 2 x 2 a cos θ where π/2 θ π/2aθ2. For integrals involvingpppxa2 x2a 2 x 2 , let x a tan θ. Thena 2 x 2 a sec θ where π/2 θ π/2pxa2 x2θa3. For integrals involvingpx2 a2½ px 2 a 2 , let x a sec θ. Thena tan θ if x awhere 0 θ π/2 a tan θ if x a where π/2 θ π

24CHAPTER 1. TECHNIQUES OF INTEGRATIONpxx2 a2θaThe restrictions on θ ensure that the function that defines the substitutionis one-to-one. In fact, these are the same intervals over which the arcsine,arctangent, and arcsecant are defined.ZExample 1.27. Find1d x.px2 9 x2Solution 1.27. First, note that none of the basic integrationp rules applies. To9 x 2 is of the hatpa 2 x 2 . So, you can use the substitution x a sin θ 3 sin θ. Using differentiation and the triangle shown below, you obtaind x 3 cos θd θ,p9 x 2 3 cos θ,x 2 9 sin2 θSo, trigonometric substitution yields3 cos θ¡ dθ9 sin2 θ (3 cos θ)ZZ1111csc2 θd θ cot θ C dθ 29 sin θ99p29 x C9x1Zx2Zdx p9 x23θZExample 1.28. Findp9 x21d x.p4x 2 1x

1.4. TRIGONOMETRIC SUBSTITUTIONSSolution 1.28. Let 2x tan θ then d x 25p1sec2 θd x and 4x 2 1 sec θ. Trigono2metric substitution producesZZZ11 sec2 θ1dθ sec θ d θdx p2sec θ24x 2 11 ln sec θ tan θ C2 1 p ln 4x 2 1 2x C2p1 4x 22xθ11ZExample 1.29. Evaluate¡x2 1 3/2 d x.³p 3¡ 3/2asSolution 1.29. Begin by writing x 2 1x 2 1 . Then, let x tan θ.pUsing d x sec2 θd θ and x 2 1 sec θ you can apply trigonometric substitution, as follows.ZZZ1sec2 θ1dθ 3 d x ¡ 3/2 d x ³psec3 θx2 1x2 1ZZ1 d θ cos θ d θ sin θ Csec θx p Cx2 1pxx2 1θ1For definite integrals, it is often convenient to determine the integrationlimits for θ that avoid converting back to x.

26CHAPTER 1. TECHNIQUES OF INTEGRATIONZExample 1.30. Evaluate2pp3x2 3d x.xpp2 3 has the form x 2 a 2 , you can consider x Solution1.30.Becausexpppp3 sec θ. Then d x 3 sec θ tan θd θ and x 2 3 3 tan θ. To pdeterminethe upper and lower limits of integration, use the substitution x 3 sec θ asfollows.pwhen x 3 sec θ 1 θ 02πwhen x 2 sec θ p θ 63So, you haveZ2p3pπ/6¡p ¡p 3 tan θ3 sec θ tan θdθp03 sec θZ π/6 p 3 tan2 θ d θ0p Z π/6 ¡ 2 sec θ 1 d θ 30pp3π 3 [tan θ θ]π/60 1 6x2 3dx xZExercise 1.4. Evaluate the following integrals.Z p1.x 1 x2 d x1dxp49 x 2Zp3. (x 1) x 2 2x 2 d xZ2.3/5 pZ4.09 25x 2 d x1dx4 4x 2 x 4Z r1 x6.dxxZ5.

1.5. PARTIAL FRACTIONS7.Z p1 e 2x d xZ8.cos xZ r9.27p4 sin2 x 9 d xx 1d x. Hint: Multiply byx 1rx 1x 11.5 Partial FractionsThis section examines a procedure for decomposing a rational function intosimpler rational functions to which you can apply the basic integration formulas. This procedure is called the method of partial fractions. Its use depends on the ability to factor the denominator, and to find the partial fractions.Recall from algebra that every polynomial with real coefficients can be factored into linear and irreducible quadratic factors. For instance, the polynomial x 5 x 4 x 1 can be written as¡ x 5 x 4 x 1 (x 1) (x 1)2 x 2 1¡ where (x 1) is a linear factor, (x 1)2 is a repeated linear factor, and x 2 1is an irreducible quadratic factor. Using this factorization, you can write thepartial fraction decomposition of the rational expression as followsP (x)x5 x4 x 1 ABCDx E 22x 1 x 1 (x 1)x 1where P (x) is a polynomial of degree less than 5, and A, B,C , D, E are constants.N(x)Note 1.2. Decomposition ofInto Partial FractionsD(x)1. Divide if improper: If N (x)/D(x) is an improper fraction (that is, if thedegree of the numerator is greater than or equal to the degree of thedenominator), divide the denominator into the numerator to obtainN (x)N (x) (a polynomial) D(x)D(x)where the degree of N (x) is less than the degree of D(x). Then applyN (x)Steps 2, 3, and 4 to the proper rational expression.D(x)

28CHAPTER 1. TECHNIQUES OF INTEGRATION2. Factor denominator: Completelyfactor¡ n the denominator into factors ofthe form (αx β)m and ax 2 bx c where ax 2 bx c is irreducible.3. Linear factors: For each factor of the form (αx β)m the partial fractiondecomposition must include the following sum of m fractions.A2AmA1 ··· (αx β) (αx β)2(αx β)m¡ n4. Quadratic factors: For each factor of the form ax 2 bx c the partialfraction decomposition must include the following sum of n fractions.B 1 x C 1¡ax 2 bx cB 2 x C 2ax 2 bx c1ZExample 1.31. Find ¡x 2 5x 6 2 · · · ¡B n x C nax 2 bx c nd x.Solution 1.31. Because x 2 5x 6 (x 3)(x 2) you should include one partial fraction for each factor and write1x 2 5x 6 AB x 3 x 2where A and B are to be determined. Multiplying this equation by the leastcommon denominator (x 3)(x 2) yields the basic equation1 A(x 2) B (x 3)Because this equation is to be true for all x, you can substitute any convenientvalues for x to obtain equations in A and B . The most convenient values arethe ones that make particular factors equal to 0. To solve for A, let x 3 toobtain A 1. To solve for B , let x 2 to obtain B 1. So, Z ·Z111dx dxx 2 5x 6x 3 x 2 x 3 C ln x 3 ln x 2 C ln x 2 ZExample 1.32. Evaluate5x 2 20x 6d x.x 3 2x 2 x

1.5. PARTIAL FRACTIONS29Solution 1.32. Because x 3 2x 2 x x(x 1)2 you should include one fractionfor each power of x and x 1 and writeBC5x 2 20x 6 A 32x 2x xx (x 1) (x 1)2Multiplying by the least common denominator x(x 1)2 yields the basic equation5x 2 20x 6 A(x 1)2 B x(x 1) C xTo solve for A let x 0. This eliminates the B and C terms and yields A 6.To solve for C let x 1. This eliminates the A and B terms and yields C 9.The most convenient choices for x have been used, so to find the value of B ,you can use any other value of x along with the calculated values of A and C .Using x 1, A 6, and C 9 produces B 1. So, it follows that Z ·Z6195x 2 20x 6dx dxx 3 2x 2 xx (x 1) (x 1)2(x 1) 1 C 6 ln x ln x 1 9 1 6 x 9 C ln x 1 x 1When using the method of partial fractions with linear factors, a convenient choice of x immediately yields a value for one of the coefficients. Withquadratic factors, a system of linear equations usually has to be solved, regardless of the choice of x.Z2x 3 4x 8 ¡ d x.Example 1.33. Find ¡ 2x x x2 4¡ ¡ ¡ Solution 1.33. Because x 2 x x 2 4 x (x 1) x 2 4 you should includeone partial fraction for each factor and writeABCx D2x 3 4x 8 ¡ 222x x 1x 4x x x 4¡ Multiplying by the least common denominator x (x 1) x 2 4 yields the basic equation¡ ¡ 2x 3 4x 8 A(x 1) x 2 4 B x x 2 4 (C x D)x(x 1)¡

30CHAPTER 1. TECHNIQUES OF INTEGRATIONTo solve for A, let x 0 and obtain A 2. To solve for B , let x 1 and obtainB 2. At this point, C and D are yet to be determined. You can find theseremaining constants by choosing two other values for x and solving the resulting system of linear equations. If x 1, then, using A 2 and B 2 youcan obtain C D 2. If x 2, you have 2C D 8. Solving these two linearequations yields C 2 and consequently D 4. It follows that Z ·Z222x42x 3 4x 8¡ ¡ dx dxx x 1 x2 4 x2 4x2 x x2 4³x ¡ 2 ln x 2 ln x 1 ln x 2 4 2 tan 1 C2An improper rational function can be integrated by performing a long division and expressing the function as the quotient plus the remainder over thedivisor. The remainder over the divisor will be a proper rational function.Z 3x x2 1d x.Example 1.34. Findx2 1Solution 1.34. The integrand is an improper rational function since the numerator has degree 3 and the denominator has degree 2. Thus, we first perform the long division.x 12x 1 32x x 13 x xx2 x 1 x2 1 x 2It follows that the integrand can be expressed asx3 x2 1x 2 x 1 22x 1x 1and hence Z 3Z ·x x2 1x 2dx x 1 2dxx2 1x 1ZZZZ12x1 x dx 1dx dx 2dx222 x 1x 1 11 ¡ x 2 x ln x 2 1 2 tan 1 x C22

1.5. PARTIAL FRACTIONS31Some times it is not necessary to use the partial fractions technique on allrational functions like in the previous example. Also, if the integrand is notin reduced form, reducing it may eliminate the need for partial fractions, asshown in the following example.Zx2 x 2d x.Example 1.35. Evaluatex 3 2x 4Solution 1.35.ZZ (x 2)x2 x 2(x 1) ¡dx x 2 2x 2

Calculus II : For Science and Engineering Lecture Notes for Calculus 102 Feras Awad Mahmoud Last Updated: October 12, 2016 1. 2 Feras Awad Mahmoud Department of Basic Sciences Philadelphia University JORDAN 19392 TEXTBOOK: This book is strongly recommended for Calculus 102 as well as a reference