Study Advice Service Mathematics Worksheet Integration 2 This is one of a series of worksheets designed to help you increase your confidence in handling Mathematics. This worksheet contains both theory and exercises which cover:1. Integration by Substitution 2. Integration by parts 3. Reduction methods There are often different ways of doing things in Mathematics and the methods suggested in the worksheets may not be the ones you were taught. If you are successful and happy with the methods you use it may not be necessary for you to change them. If you have problems or need help in any part of the work then there are a number of ways you can get help. For students at the University of Hull Ask your lecturers Contact the Study Advice Service in the Brynmor Jones Library where you can access the MathematicsTutor, or contact us by email. Come to a Drop-In session organised for your department Look at one of the many textbooks in the library. For others Ask your lecturers Access your Study Advice or Maths Help Service Use any other facilities that may be available. If you do find anything you may think is incorrect (in the text or answers) or want further help please contact us by email. Tel: 01482 466199 Web: www.hull.ac.uk/studyadvice Email: studyadvice@hull.ac.uk 1. Integration by substitution (or change of variable) Find the integral of ax b n ie. find I ax b dx by the Fundamental Theorem of calculus this means that dI ax b n dx du dI a and ax b n u n Put u ax b then dx dx n By the Chain Rule Giving hence integrating dI dI du dx du dx dI un a du dI 1 n u du a 1 I u n du a 1 u n 1 ax b n 1 c c a n 1 an 1 In practice, when doing the integration, we simplify this to I = ax bn dx ) u ax b gives du a dx (i.e. dx du a substituting giving, I = n u du 1 n ax bn 1 c 1 u n 1 u du = c a a a n 1 an 1 Examples 1. By substituting u tan x, evaluate I tan5 x sec2 x dx du sec 2 x du sec 2 xdx Substituting u tan x gives dx So tan5 x sec 2 x dx tan5 x sec 2 x dx u6 u du c 6 5 1 tan6 6 x c 1 2. By substituting u 3x 4 6 , evaluate the integral I Put u 3x 4 6 then du 12 x 3 du 12 x 3 dx dx x3 (3 x 4 6)5 dx 1 12 x 3 1 12 x 3 dx dx 12 (3x 4 6) 5 12 (3x 4 6) 5 1 du 1 5 we have I 5 u du 12 u 12 Writing I as 1 u 4 1 c 12 4 48u 4 1 c 48(3 x 4 6)4 c 3 3. Evaluate 04 tan 5 x sec 2 x dx In example 1 we found, by substituting u tan x , tan 5 x sec 2 x dx = 16 tan 6 x c Hence 3 4 0 5 2 tan x sec x dx 1 tan 6 6 xc 4 6 3 4 0 1 tan 6 3 c 1 tan 6 0 c 1 16 c 0 c 1 6 6 6 Notice that the value of c is irrelevant as it cancels out. This is why we do not need to put it in when doing definite integrals. In practice it is also easier to change the limits when we do the substitution. I= 3 4 0 tan 5 x sec 2 x dx . du sec 2 x du sec 2 xdx dx the limits become tU tan 3 1, t L tan0 0 4 Substituting u tan x gives So 3 1 6 I 0 4 tan5 x sec 2 x dx 01 t 5 dt 61 t 6 61 1 0 61 0 Exercise 1 Find the following integrals using the given substitution dx 1. x 3x 2 1 dx, u 3x 2 1 2. u 4 3x 3 4 3 x 4 3. 5 x 3 x 4 3 dx, u x 4 3 4. sin3 x dx, u cos x e 3x 6. x 3x 1 dx, u 3x 1 5. dx, u e 3 x 3x e 2 2 By using an appropriate substitution find the following integrals. 7. 10. 13. t2 te dt 8. x a 2 x 2 dx x 1 x2 cos v 1 sin v dv e 3 x 1 11. 14. 2 3 1 3x 5 x x dx dx 1 e 3x dx 5 p5 1 0 9. 12. 2 3 p6 t 3 3t 2 5 dp dt 15. 5 02 cos x sin x dx 2 Integration by parts It is not possible to integrate the function xe x by any of the methods outlined 2 previously. The more complicated looking function xex can be done by substituting u x 2 but xe x can only be integrated as a product. du etc., differentiating a product ( uv ) gives the rule dx d uv u ' v uv' dx uv u ' v dx uv' dx Integrating both sides we get Writing u ' for which gives the basic rule for integrating a product uv' dx uv u ' v dx On the left hand side we have u and v ' and we need u ' and v on the right hand side. This implies that we must be able to integrate one of the functions (easily) and differentiate the other (always possible). In some cases there is no problem in deciding which is which. For instance if one of the functions is ln x we need to differentiate it as ln x cannot be integrated at sight. If both functions can be integrated you should be able to decide which is which with experience. If your first choice doesn’t work try again!! Examples 1. Evaluate xe x dx Both x and e x can be integrated so which do we choose? First attempt: Take u x and v' e x then u' 1 and v e x Using uv' dx uv u ' v dx we get xe x dx xex 1 e x dx xex e x c 3 Second attempt: Take u e x and v ' 1x then u ' e x and v = 12 x 2 Using uv' dx uv u ' v dx we get xe x dx e x x e x dx 1 2 2 1 x 2e x 2 x 1 2 2 21 x2e x dx The last part x 2e x dx is ‘worse’ than we started with as it involves the product of x 2 and e x whereas in the first attempt we could do the final integral immediately, in the second attempt we’d have to try using integration by parts again (in practice you’d come back to where you started!) In general if both functions can be integrated easily and you have a power of x , then differentiate the x -term. Function x sin x dx t Differentiate x 2 t 1 2. Evaluate sin x t2 e dt x tan x Integrate tan 1 x dx x et x (as you can’t integrate tan1 x directly) 2 ln x dx Both terms can be differentiated but ln x cannot be integrated immediately 1 x3 Take u ln x and v' x 2 then u ' and v x 3 Using uv' dx uv u ' v dx and writing the function in the order uv' gives x 2 ln x x dx ln x 3 3 3 1 x dx x 3 x 3 ln x 1 2 x 3 ln x x 3 x dx c 3 3 3 9 1 x 3 3 ln x 1 c 1 x 3 ln x 3 1 c 9 9 Notice that the answer can be written in a number of different forms, all of which are correct. Sometimes the question dictates which form to leave the answer in. 4 3. Evaluate 0 x sin x dx Take u x and v' sin x then u' 1 and v cos x 0 x sin x dx x cos x 0 0 1 cos x dx x cos x sin x 0 cos sin 0 1 4. Evaluate x 2 3 x e dx Take u x 2 and v ' e 3 x Giving I = x 2 3 x e then u' 2 x and v 13 e 3 x dx x 2 1 e3x 2 x 1 e3x dx 3 3 2 3 x 3 x 1 x e 2 xe dx 3 3 3 x we now need to integrate xe dx by parts It is important that this is not done separately but everything is kept together otherwise some parts may get lost. Take u x and v ' e 3 x then u' 1 and v 13 e 3 x We get x 2 3 x e dx 13 x 2 e 3 x 23 xe3 x dx x 1 e 3 x 1 e 3 x c 3 9 13 x 2 e 3 x 23 x 13 e 3 x 13 e 3 x dx 13 x 2 e 3 x 23 2 e 3 x c 13 x 2 e 3 x 92 xe 3 x 27 1 e 3 x 9 x 2 6 x 2 c 27 Exercise 2 Use the method of integration by parts to find the following 2. x lnx 2 dx 1. x sin x dx 3. u 2 sin u du ln t 4. s 2 ln s ds 6. sin 1 w dw 5. dt 3 t 7. y 2 e 3 y dy 9. x x 1 dx 8. xx 14 dx 10. 13. 1 0 4 2 xx 15 dx t 1ln2t dt 11. 6 3 x x2 dx 12. 14. 0 4 u cos 2u du 15. 2 3 s s e 0 4 ln 1 2 ds d 5 3. Reduction methods cos n x dx can be evaluated, fairly easily, for some values of n, for instance. a) cos 2 x dx . From cos 2 x 2 cos2 x 1 we have cos2 x 1 cos 2 x 1 2 hence cos 2 x dx 12 cos 2 x 1 dx 12 12 sin 2 x x c b) cos 3 x dx 1 sin2 x cos x dx using cos 2 x 1 sin 2 x Substituting s sin x, ds cos x dx giving 3 2 2 3 3 cos x dx 1 sin x cos x dx 1 s ds s 13 s c sin x 13 sin x c c) cos 4 x dx 12 1 cos 2 x dx = 41 1 2 cos 2 x cos 1 1 2 cos 2 x 1 1 cos 4 x dx 4 2 1 x sin 2 x 1 x 1 sin 4 x c 4 8 4 2 2 2 x dx What about cos 20 x dx or cos31 x dx ? In cases like this a reduction method, based on integration by parts, is useful but not necessarily the only method. Examples 1. If I n = cos n x dx (a) show that I n 1 n 1 sin x cos n 1 x In2 n n (b) Find the values of (i) I 5 (ii) I 4 (a) Writing cos n x as cos x cos n1 x we have I n cos x cos n1 x dx Using uv' dx uv u ' v dx and taking u cos n1 x and v' cos x then u' n 1 cos n2 x sin x and v sin x then I n cos n 1 x sin x ( n 1) cos n 2 x ( sin x ) sin x dx cos n 1 x sin x ( n 1) cos n 2 x sin 2 x dx cos n 1 x sin x ( n 1) cos n 2 x (1 cos 2 x ) dx cos n 1 x sin x ( n 1) cos n 2 x cos n x dx cos n 1 x sin x ( n 1) cos n 2 x dx ( n 1) cos n x dx I n cos n 1 x sin x ( n 1)I n 2 ( n 1)I n I n ( n 1)I n nI n cos n 1 x sin x ( n 1)I n 2 1 n 1 I n sin x cos n 1 x I n2 n n 6 (b) (i) From the formula I 5 1 sin x cos 4 x 4 I 3 5 5 but I 3 1 sin x cos 2 x 2 I 1 and I 1 cos x dx sin x C 3 3 hence I 5 1 sin x cos 4 x 4 1 sin x cos 2 x 2 sin x C 5 5 3 3 1 sin x cos 4 x 4 sin x cos 2 x 8 sin x K 5 15 15 (b) (ii) From the formula I 4 1 sin x cos 3 x 3 I 2 4 4 but I 2 1 sin x cos x 1 I 0 and I 0 2 2 0 cos x dx 1 dx x C hence I 4 1 sin x cos 3 x 3 1 sin x cos x 1 x C 4 4 2 2 1 sin x cos 3 x 3 sin x cos x 3 x K 4 8 2. Find a reduction formula for x 8 n x e dx taking u x n and v e x then u = nx n 1 and v e x we get I n x ne x nx n 1e x dx x ne x n x n 1e x dx x ne x nI n 1 Hence reduction formula is I n x ne x nI n 1 To evaluate x e 5 x dx we have I 5 x5e x 5I 4 I 4 x 4 e x 4I 3 I3 x3e x 3I 2 I 2 x 2e x 2I1 I1 xex I 0 I 0 e x dx e x c Combining these (very carefully!!) I1 xex I 0 xex e x C I 2 x2e x 2I1 x2e x 2( xex e x C ) x2 2 x 2 e x 2C I3 x3e x 3I 2 x3e x 3 x 2 2 x 2 e x 2C x3 3x 2 6 x 6 e x 6C I 4 x 4 e x 4 I 3 x 4 e x 4 x 3 3x 2 6 x 6 e x 6C x 4 4 x 3 12 x 2 24 x 24 e x 24C I 5 x 5 e x 5 I 4 x 5 e x 5 x 4 4 x 3 12 x 2 24 x 24 e x 24C x 5 5 x 4 20 x 3 60 x 2 120 x 120 e x K 7 Hence x e 5 x dx x5 5 x 4 20 x3 60 x 2 120 x 120 e x K Exercise 3 Establish reduction formulas for the following 1. 2. sin x dx hence find sin x dx, sin n 3 xln x dx hence find xln x dx 5 n 4 x dx ANSWERS Exercise 1 1. 3x 2 1 9 3 2 2. c 4. 1 cos 3 x cos x c 3 2 7. 1 e t c 2 10. 1 a 2 x 2 3 3 2 13. 1 x 2 c c 1 64 x 32 c 3. x 4 35 c 4 5 3 5. 1 ln e 3 x 2 c 6. 2 (3x 1) 2 2 (3x 1) 2 c 8. ln1 sin v c 9. 1 ln 4 3 45 6 27 3 7 ln 1 e 3 x 11. c 3e 12. 1 ln 22 6 14. 1 5 x x 3 c 15. 1 6 6 6 8 Exercise 2 2 3. 2 u 2 cos u 2u sin u c 5. 1 2 ln t 4t 2 2. 1 x 2 4 lnx 2 1 x 2 4 x c 1. sin x x cos x c 4 4. 1 s 3 ln s 1 c 3 3 6. w sin 1 w 1 w 2 c c y2 2y 2 9y2 6y 2 c 7. e 3 y c e3y 3 9 27 27 8. x 15 5 x 1 c 2x 1 2 3x 2 9. c 15 3 30 10. 1 11. 26 42 3 12. 2e 2 6 Exercise 3 n 1 1. I n I n 2 1 cos x sin n1 x n n 13. 4 ln 8 1 I 5 1 cos x 3 sin 4 x 4 sin3 x 8 c 15 14. 2 8 15. 3 2 ln 2 4 I 4 1 2 sin3 x cos x 3 sin x cos x 3x c 8 2. I n 1 x 2 ln x n nI n1 2 I 3 1 x 2 4ln x 3 6ln x 2 6 ln x 3 c 8 We would appreciate your comments on this worksheet, especially if you’ve found any errors, so that we can improve it for future use. Please contact the Maths tutor by email at studyadvice@hull.ac.uk Updated 29th November 2004 The information in this leaflet can be made available in an alternative format on request. Telephone 01482 466199 © 2009 9