1. How to integrate some special products of two functions
Integration by parts
How to integrate some special products of two functions
April 2011
纪光 - 北京 景山学校

2. Problem
Not all functions can be integrated by using simple derivative formulas backwards …
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 2011
纪光 - 北京 景山学校

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)
hence
April 2011
纪光 - 北京 景山学校

4. Examples of integrals products of composite functions
April 2011
纪光 - 北京 景山学校

5. Formulas of integrals of products made of composite functions
April 2011
纪光 - 北京 景山学校

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 2011
纪光 - 北京 景山学校

7. Example 1 Integration by parts u’ =(x)’ = 1 u = x sin x = (- cos x)’
u = 1st part
(to derive)
v’ = 2nd part
(to integrate)
u’ =(x)’ = 1
u = x
sin x = (- cos x)’
v’ = sin x
u’v = 1.(- cos x)
v = - cos x
April 2011
纪光 - 北京 景山学校

8. Example 1 Integration by parts u = x  u’ = 1 v’ = sin x  v = - cos x
April 2011
纪光 - 北京 景山学校

9. Example 2 Integration by parts u’ =(x2)’ = 2x u = x2 cos x = (sin x)’
1st part
(to derive)
2nd part
(to integrate)
u’ =(x2)’ = 2x
u = x2
cos x = (sin x)’
v’ = cos x
u’v = 2x.sin x
v = sin x
April 2011
纪光 - 北京 景山学校

10. Example 2 Integration by parts u = x2  u’ = 2x v’ = cos x  v = sin x
April 2011
纪光 - 北京 景山学校

11. Example 3 Integration by parts u’ =(ln x)’ = u = ln x x = =>v = .
1st part
(to derive)
2nd part
(to integrate)
u’ =(ln x)’ =
u = ln x
x = =>v = .
v’ = x
u’v =
April 2011
纪光 - 北京 景山学校

12. Example 3 Integration by parts u = ln x  u’ = v’ = x  v = April 2011
纪光 - 北京 景山学校

13. Example 4 Integration by parts u’ =(ln x)’ = u = ln x
1st part
(to derive)
2nd part
(to integrate)
u’ =(ln x)’ =
u = ln x
1 = (x)’=> v =
v’ = 1
u’v = 1
April 2011
纪光 - 北京 景山学校

14. Example 4 Integration by parts u = ln x  u’ = v’ = 1  v = x
April 2011
纪光 - 北京 景山学校

15. Example 5 Integration by parts u’ =(ln x)’ = u = ln x
1st part
(to derive)
2nd 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 2011
纪光 - 北京 景山学校

16. Example 5 Integration by parts u = ln x  u’ =
v’ = ln x  v = x.ln x - x
April 2011
纪光 - 北京 景山学校

17. Problem I Integration by parts Use the IBP formula to calculate :
exp(exp(1) + 1) - exp(exp(1))
April 2011
纪光 - 北京 景山学校

18. Problem II Integration by parts Use the IBP formula to calculate :
April 2011
纪光 - 北京 景山学校

19. Problem III Integration by parts Let , for any Integer n ≥ 0 :
Use the IBP formula to prove that :
Then find I0, I1, I2
April 2011
纪光 - 北京 景山学校

20. Problem IV Integration by parts Use the IBP formula to prove that :
Let , for any Integer n ≥ 0 : [Wallis Integral]
Problem IV
Use the IBP formula to prove that :
Calculate I0 and I1
Find a short formula for I2n
Find a short formula for I2n+1
April 2011
纪光 - 北京 景山学校

21. Problem V Integration by parts Use twice the IBP formula to calculate
April 2011
纪光 - 北京 景山学校

22. Problem VI Integration by parts Use twice the IBP formula to calculate
Anwser : I =
April 2011
纪光 - 北京 景山学校

23. Problem VII Integration by parts Calculate I + J
Use IBP to calculate I – J
Find I and J
Anwser : I =
April 2011
纪光 - 北京 景山学校

24. Problem VIII Integration by parts Calculate I1
Find a relationship between In and In (n ≥ 1)
Show that
Anwser : I =
April 2011
纪光 - 北京 景山学校