April 2009 纪光 - 北京 景山学校 Integration by parts – p.1 How to integrate some special products of two functions

1. Not all functions can be integrated by using simple derivative formulas backwards … 2. Some functions look like simple products but cannot be integrated directly. Ex. f(x) = x.sin x but … u’.v’≠ (u.v)’ So what ?!? April 2009 纪光 - 北京 景山学校 2

April 2009 纪光 - 北京 景山学校 Integration by parts – p.3 Reminder 1 Derivative of composite functions if g(x) = f[u(x)] and u and f have derivatives, then g’(x) = f’[u(x)]. u’(x) hense

April 2009 纪光 - 北京 景山学校 Integration by parts – p.4 Examples of integrals products of composite functions

April 2009 纪光 - 北京 景山学校 Integration by parts – p.5 Formulas of integrals of products made of composite functions

April 2009 纪光 - 北京 景山学校 Integration by parts – p.6 Reminder 2 Derivative of the Product of 2 functions If u and v have derivatives u’ and v’, then (u.v)’ = u’.v + u.v’ u.v’ = (u.v)’ - u’.v hence by integration of both sides

April 2009 纪光 - 北京 景山学校 Integration by parts – p.7 Example 1 u = 1 st part (to derive) u = 1 st part (to derive) v’ = 2 nd part (to integrate) u’ =(x) ’ = 1 u = x sin x = (- cos x ) ’ v’ = sin x u’v = 1.(- cos x) v = - cos x

April 2009 纪光 - 北京 景山学校 Integration by parts – p.8 Example 1 u = x  u’ = 1 v’ = sin x  v = - cos x

April 2009 纪光 - 北京 景山学校 Integration by parts – p.9 Example 2 1 st part (to derive) 1 st part (to derive) 2 nd part (to integrate) u’ =(x 2 ) ’ = 2 x u = x 2 cos x = (sin x ) ’ v’ = cos x u’v = 2 x.sin x v = sin x

April 2009 纪光 - 北京 景山学校 Integration by parts – p.10 Example 2 u = x 2  u’ = 2x v’ = cos x  v = sin x

April 2009 纪光 - 北京 景山学校 Integration by parts – p.11 Example 3 1 st part (to derive) 1 st part (to derive) 2 nd part (to integrate) u’ =(ln x) ’ = u = ln x u’v = x = =>v =. v’ = x

April 2009 纪光 - 北京 景山学校 Integration by parts – p.12 Example 3 u = ln x  u’ = v’ = x  v =

April 2009 纪光 - 北京 景山学校 Integration by parts – p.13 Example 4 1 st part (to derive) 1 st part (to derive) 2 nd part (to integrate) u’ =(ln x) ’ = u = ln x 1 = (x)’=> v = x v’ = 1 u’v = 1

April 2009 纪光 - 北京 景山学校 Integration by parts – p.14 Example 4 u = ln x  u’ = v’ = 1  v = x

April 2009 纪光 - 北京 景山学校 Integration by parts – p.15 Example 5 1 st part (to derive) 1 st part (to derive) 2 nd part (to integrate) u’ =(ln x) ’ = u = ln x ln x = (x ln x – x)’ v’ = ln x u’v = ln x - 1 v = x ln x – x

April 2009 纪光 - 北京 景山学校 Integration by parts – p.16 Example 5 u = ln x  u’ = v’ = ln x  v = x. ln x - x

April 2009 纪光 - 北京 景山学校 Integration by parts – p.17 Problem I Use the IBP formula to calculate :

April 2009 纪光 - 北京 景山学校 Integration by parts – p.18 Problem II Use the IBP formula to calculate :

April 2009 纪光 - 北京 景山学校 Integration by parts – p.19 Problem III Use the IBP formula to prove that : Let, for any n > 0 : Then find I 2, I 3, I 4

April 2009 纪光 - 北京 景山学校 Integration by parts – p.20 祝好云 谢谢 再见