Solved exercises simple integration
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This document contains 7 exercises involving calculating integrals using techniques like integration by parts, trigonometric substitutions, and partial fraction decomposition. It also contains solutions to each exercise that demonstrate the step-by-step working to evaluate the integrals. The document is from a university course on simple integration and contains examples commonly used to teach integration techniques.
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Solved exercises simple integration
- 1. Lebanese University - Faculty of Sciences Section ¶ Chapter 1: Simple Integration Solved Problems Dr. Kamel ATTAR [email protected] F Wednesday 24/Mars/2021 F
- 2. 2Ú20 Exercises Solutions 1 Exercises 2 Solutions Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 3. 3Ú20 Exercises Solutions . Exercise 1. Calculate the following integrals: a) Z sin2 (x) dx b) Z cosh2 (x) dx c) Z tan2 (x) dx . Go to Solution . Exercise 2. Evaluate by using change of variable the following integrals : a) Z (x3 + x)5 (3x2 + 1) dx b) Z p 2x + 1 dx c) Z 2x dx 3 p x2 + 1 d) Z arccos(x) − x p 1 − x2 dx e) Z sin(2x)esin2 x dx . Go to Solution . Exercise 3. Evaluate I(x) = Z x 2 − x2 + 2x dx then deduce J(x) = Z dx 1 + x + 2 √ 1 − x Go to Solution . Exercise 4. Calculate using the integration by parts the following integrals: a) Z x sin 2xdx b) Z x ln x dx c) Z x2 ln(x)dx d) Z (x2 + 7x − 5) cos(2x)dx e) Z x2 e3x dx f) Z x arcsin x p 1 − x2 dx g) Z arcsin2 x 2 p 4 − x2 dx h) Z x ln 1 + x 1 − x dx . Go to Solution Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 4. 4Ú20 Exercises Solutions . Exercise 5. Using twice integration by parts, evaluate: Z e−x cos(x)dx . Go to Solution . Exercise 6. Calculate: a) Z 2x2 x4 − 1 dx b) Z dx x2(x − 1)3 c) Z 1 (x2 + 1)2 dx Go to Solution . Exercise 7. Calculate: a) Z dx x(x2 + 2x + 5) b) Z dx x(x2 + 1)2 c) Z dx x(x5 + 1)2 . Go to Solution Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 5. 5Ú20 Exercises Solutions Solution 1. a) We have that the power of sin is even, then we can write the integral as follows: Z sin2 (x) dx = Z 1 − cos(2x) 2 dx = x 2 − sin(2x) 4 + C. b) Same for the hyperbolic function Z cosh2 (x) dx = Z 1 + cosh(2x) 2 dx = x 2 + sinh(2x) 4 + C. c) We have tan2 (x) = sin x cos x 2 = 1 − cos2 x cos2 x = 1 cos2 x − 1 . Then Z tan2 (x) dx = Z 1 cos2 x dx − Z dx = tan x − x + C. Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 6. 6Ú20 Exercises Solutions Solution 2. a) We set u = x3 + x. Then du = (3x2 + 1)dx , so that by substitution we have Z (x3 + x)5 (3x2 + 1) dx = Z u5 du = u6 6 + C = (x3 + x)6 6 + C. b) We set u = 2x + 1. Then du = 2dx , so that by substitution we have Z p 2x + 1 dx = Z p 2x + 1 dx = 1 2 Z u 1 2 u0 = 1 3 u3/2 + C = 1 3 (2x + 1)3/2 + C . c) Substitute u = x2 + 1 then u0 = 2x and so Z 2x dx 3 p x2 + 1 = Z u0 u1/3 = Z u−1/3 u0 = u2/3 2/3 + C = 3 2 u2/3 + C = 3 2 (x2 + 1)2/3 + C. d) Z arccos(x) − x p 1 − x2 dx = Z arccos(x) p 1 − x2 dx + Z −x p 1 − x2 dx We set u = arccos x and v = 1 − x2 then du = − 1 p 1 − x2 dx and dv = −2xdx. Thus Z arccos(x) − x p 1 − x2 dx = − Z u du + Z dv 2 √ v = − arccos2 x 2 + p 1 − x2 + C . Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 7. 7Ú20 Exercises Solutions e) We havesin(2x) = 2 cos(x) sin(x), we obtain Z sin(2x)esin2 x dx = Z 2 cos x sin x esin2 x dx Using change of variable t = sin2 x, we get dt = 2 cos x sin x dx. Thus Z sin(2x)esin2 x dx = Z et dt = et + C = esin2 x + C Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 8. 8Ú20 Exercises Solutions Solution 3. a) I(x) = Z xdx −x2 + 2x + 2 = − 1 2 Z −2xdx −x2 + 2x + 2 u = −x2 + 2x + 2 = − 1 2 Z −2x + 2 −x2 + 2x + 2 dx + Z 1 −x2 + 2x + 2 dx = − 1 2 Z u0 u + Z 1 √ 3 2 − (x − 1)2 dx u = x − 1 u0 = 1 , a = √ 3 = − 1 2 ln |u| + Z u0 √ 3 2 − u2 = − 1 2 ln | − x2 + 2x + 2| + 1 √ 3 arg sh x − 1 √ 3 + C . b) We set t = p 1 − x then x + 1 = 2 − t2 and dx = −2tdt. Hence J(x) = Z −2t dt 2 − t2 + 2t = −2I(t) = ln | − t2 + 2t + 2| − 2 √ 3 arg sh t − 1 √ 3 + C Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 9. 9Ú20 Exercises Solutions Solution 4. a) We need to choose u. In this question we don’t have any of the functions suggested in the priorities list above. We could let u = x or u = sin 2x, but usually only one of them will work. In general, we choose the one that allows to be of a simpler form than u. So, we choose u = x and so v0 will be the rest of the integral, v0 = sin 2x. We have u = x × v0 = sin 2x u0 = 1 ← − − R v = − 1 2 cos 2x Substituting these 4 expressions into the integration by parts formula, we get Z x sin 2xdx = x × − 1 2 cos 2x − Z − 1 2 cos 2x × 1dx = − 1 2 x cos 2x + 1 2 Z cos 2x dx = − 1 2 x cos 2x + 1 4 sin 2x . b) u = ln x × v0 = x u0 = 1 x ← − − R v = x2 2 Then, applying the formula Z x ln x dx = x2 2 ln x − Z x2 2 · 1 x dx = x2 2 ln x − x2 4 + C Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 10. 10Ú20 Exercises Solutions c) u = ln x × v0 = x2 u0 = 1 x ← − − R v = x3 3 Then Z x2 ln x dx = x3 3 ln x − 1 3 Z x2 dx = x3 3 ln x − x3 9 + C. d) u x2 + 7x − 5 cos(2x) v0 + u0 2x + 7 1 2 sin(2x) v − u00 2 − 1 4 cos(2x) Z v + u000 0 − 1 8 sin(2x) Z Z v Z (x2 + 7x − 5) cos(2x) = (x2 + 7x − 5) 1 2 sin(2x) − (2x + 7) − 1 4 cos(2x) + (2) = 1 2 x2 + 7x − 11 2 sin(2x) + 1 2 x + 7 2 cos(2x) + C Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 11. 11Ú20 Exercises Solutions e) We have to make a choice and let one of the functions in the product equal u and one equal to v0 . As a general rule we let u be the function which will become simpler when we differentiate it. In this case it makes sense to let u = x2 × v0 = e3x u0 = 2x ← − − R v = 1 3 e3x Then, using the formula for integration by parts, Z x2 e3x dx = 1 3 e3x · x2 − Z 1 3 e3x · 2x dx = 1 3 x2 e3x − 2 3 Z xe3x dx . The resulting integral is still a product. It is a product of the functions 2 3 x and e3x . We can use the formula again. This time we choose u = 2 3 x × v0 = e3x u0 = 2 3 ← − − R v = 1 3 e3x So Z x2 e3x dx = 1 3 x2 e3x − 2 3 Z xe3x dx = 1 3 x2 e3x − 2 3 x · 1 3 e3x − Z 1 3 e3x · 2 3 dx = 1 3 x2 e3x − 2 9 xe3x + 2 27 e3x + C. Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 12. 12Ú20 Exercises Solutions f) u = arcsin(x) × v0 = x p 1 − x2 u0 = 1 p 1 − x2 ← − − R v = − p 1 − x2 Then Z x arcsin x p 1 − x2 dx = − p 1 − x2 arcsin x + x + C g) Let I(x) = Z arcsin2 x 2 p 4 − x2 dx. Using integration by parts u = arcsin x 2 × v0 = arcsin x 2 1 2 q 1 − x 2 2 u0 = 1 2 q 1 − x 2 2 ← − − R v = 1 2 arcsin2 x 2 Thus I(x) = 1 2 arcsin3 x 2 − 1 4 I(x) = ⇒ I(x) = 2 5 arcsin3 x 2 + C . Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 13. 13Ú20 Exercises Solutions h) We set u = ln 1 + x 1 − x × v0 = x u0 = 2 1 − x2 ← − − R v = x2 2 We obtain Z x ln 1 + x 1 − x dx = x2 2 ln 1 + x 1 − x + Z x2 1 − x2 dx = x2 2 ln 1 + x 1 − x − Z 1 − 1 − x2 1 − x2 dx = x2 2 ln 1 + x 1 − x − x + Z 1 1 − x2 dx = x2 2 ln 1 + x 1 − x − x + 1 2 ln
- 17. 1 + x 1 − x
- 21. + C Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 22. 14Ú20 Exercises Solutions Solution 5. I(x) = Z e−x cos(x) dx. We set u = e−x × v0 = cos(x) u0 = −e−x ← − − R v = sin(x) Then I(x) = e−x sin(x) + Z e−x sin(x) dx . Same for Z e−x sin(x) dx we set u = e−x × v0 = sin(x) u0 = −e−x ← − − R v = − cos(x) we obtain Z e−x sin(x) dx = −e−x cos(x) − Z e−x cos(x) = −e−x cos(x) − I(x) Thus I(x) = e−x sin(x) − e−x cos(x) − I(x) + C Finally I(x) = 1 2 sin(x) − cos(x) e−x + C Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 23. 15Ú20 Exercises Solutions Solution 6. a) The partial fraction decomposition has the form 2x2 x4 − 1 = 2x2 (x − 1)(x + 1)(x2 + 1) = A x − 1 + D x + 1 + Bx + C x2 + 1 . We clear fractions and get 2x2 = A(x + 1)(x2 + 1) + D(x − 1)(x2 + 1) + (Bx + C)(x − 1)(x + 1) . For x = 1 we obtain, A = 1 2 and for x = −1 we obtain, D = − 1 2 . Then, 2x2 (x − 1)(x + 1)(x2 + 1) = 1/2 x − 1 + −1/2 x + 1 + Bx + C x2 + 1 . For x = 0 and x = 2, we have C = 1 and B = 0. Thus 2x2 (x − 1)(x + 1)(x2 + 1) = 1/2 x − 1 + −1/2 x + 1 + 1 x2 + 1 . Finally, Z 2x2 x4 − 1 dx = Z 1/2 x − 1 dx + Z −1/2 x + 1 dx + Z 1 x2 + 1 dx = 1 2 ln |x − 1| − 1 2 ln |x + 1| + arctan(x) + C . Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 24. 16Ú20 Exercises Solutions b) Let g(x) = 1 x2(x − 1)3 . The partial fraction decomposition has the form g(x) = A x2 + B x + C (x − 1)3 + D (x − 1)2 + E x − 1 A = lim x→0 x2 g(x) = lim x→0 1 (x − 1)3 = −1 B = lim x→0 h x2 g(x) i0 = lim x→0 − 3 (x − 1)4 = −3 C = lim x→1 (x − 1)3 g(x) = lim x→1 1 x2 = 1 D = lim x→1 h (x − 1)3 g(x) i0 = lim x→1 − 2 x3 = −2 E = lim x→1 1 2 h (x − 1)3 g(x) i00 = 1 2 lim x→1 6 2x4 = 3 Thus 1 x2(x − 1)3 = − 1 x2 − 3 x + 1 (x − 1)3 − 2 (x − 1)2 + 3 x − 1 . Finally Z 1 x2(x − 1)3 dx = 1 x − 3 ln |x| − 1 2 1 (x − 1)2 + 2 x − 1 − 3 ln |x − 1| + C . Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 25. 17Ú20 Exercises Solutions c) We have Z 1 (1 + x2)2 dx = Z 1 1 + x2 dx 1 + x2 = Z 1 1 + x2 (arctan x)0 Set t = arctan x then dt = dx 1 + x2 and so Z 1 (1 + x2)2 dx = Z dt 1 + tan2(t) = Z cos2 (t) dt = t 2 + sin(2t) 4 + C = t 2 + tan t 2 1 + tan2(t) + C = arctan x 2 + x 2(1 + x2) + C Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 26. 18Ú20 Exercises Solutions Solution 7. a) The partial fraction decomposition has the form 1 x(x2 + 2x + 5) = A x + Bx + C x2 + 2x + 5 . We clear fractions and get 1 = A(x2 + 2x + 5) + (Bx + C)x . For x = 0, we obtain A = 1 5 and for x = 1 and x = −1 we get B = − 1 5 and C = − 2 5 . Then 1 x(x2 + 2x + 5) = 1 5x − x + 1 5(x2 + 2x + 5) = 1 5x − 2x + 2 10(x2 + 2x + 5) − 1 5 (x + 1)2 + 4 . Thus Z dx x(x2 + 2x + 5) = 1 5 ln |x| − 1 10 ln(x2 + 2x + 5) − 1 10 arctan x + 1 2 + C . b) Set t = 1 + x2 , then dt = 2xdx and x2 = t − 1 we obtain Z dx x(x2 + 1)2 = Z xdx x2(x2 + 1)2 = Z dt (t − 1)t2 = − 1 2 Z 1 t2 + 1 t + 1 1 − t dt = 1 2t − l Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems
- 27. 19Ú20 Exercises Solutions c) Let t = x5 + 1, we have Z 1 x(x5 + 1)2 dx = 1 5 Z dt (t − 1)t2 = A t − 1 + B t + C t2 By comparison, we find A = 1 5 , C = − 1 5 and B = 1 5 Hence 1 5 Z dt (t − 1)t2 = Z 1/5 t − 1 dt + Z 1/5 t dt − Z 1/5 t2 dt and so 1 5 Z dt (t − 1)t2 = 1 5 ln |t − 1| + 1 5 ln |t| + 1 5t + C . Go Back Dr. Kamel ATTAR | Chapter 1: Simple Integration | Solved Problems