3. Example 2

4. Example 2
Let’s find the integral:

5. Example 2
Let’s find the integral:
x sin xdx

6. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:

7. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx

8. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:

9. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x

10. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x

11. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1

12. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x

13. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:

14. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx

15. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
u(x)
x sin xdx

16. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x$$$Xv (x)
sin xdx

17. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx =

18. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx = x

19. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx = x(− cos x)

20. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx =
u(x)
x(− cos x)

21. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx = x$$$$$Xv(x)
(− cos x)

22. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx = x(− cos x)−

23. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx = x(− cos x) − 1. (− cos x) dx

24. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx = x(− cos x) − ¡¡!
u (x)
1. (− cos x) dx

25. Example 2
Let’s find the integral:
x sin xdx
Let’s write again the formula of integration by parts:
u(x)v (x)dx = u(x)v(x) − u (x)v(x)dx
In this case we choose:
u(x) = x v (x) = sin x
u (x) = 1 v(x) = − cos x
So, we have that:
x sin xdx = x(− cos x) − 1.$$$$$Xv(x)
(− cos x)dx

26. Example 2

27. Example 2
So our integral becomes:

28. Example 2
So our integral becomes:
x sin xdx = −x cos x + cos xdx

29. Example 2
So our integral becomes:
x sin xdx = −x cos x + cos xdx
And that’s easy to solve:

30. Example 2
So our integral becomes:
x sin xdx = −x cos x + cos xdx
And that’s easy to solve:
x sin xdx = −x cos x + sin x

31. Example 2
So our integral becomes:
x sin xdx = −x cos x + cos xdx
And that’s easy to solve:
x sin xdx = −x cos x + sin x

32. Trick Number 1

33. Trick Number 1
Take v (x) = 1.

34. Trick Number 1
Take v (x) = 1. For example:

35. Trick Number 1
Take v (x) = 1. For example:
ln xdx

36. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:

37. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x

38. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1

39. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x

40. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x

41. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:

42. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx

43. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx =

44. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.1dx

45. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ¨¨¨B
u(x)
ln x.1dx

46. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx

47. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x

48. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x

49. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ¨
¨¨B
u(x)
ln x.x

50. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.
v(x)
x

51. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x−

52. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x −
1
x
.xdx

53. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x −
£
£
£#
u (x)
1
x
.xdx

54. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x −
1
x
.
v(x)
xdx

55. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x −
1
x
.xdx

56. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x −
1
x
.xdx

57. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x −
1
x
.xdx = x ln x − dx

58. Trick Number 1
Take v (x) = 1. For example:
ln xdx
We take:
u(x) = ln x v (x) = 1
u (x) = 1
x v(x) = x
So we have:
ln xdx = ln x.¡¡!
v (x)
1dx
= ln x.x −
1
x
.xdx = x ln x − dx = x ln x − x

59. Trick Number 1

60. Trick Number 1
So we have:

61. Trick Number 1
So we have:
ln xdx = x (ln x − 1)

62. Trick Number 1
So we have:
ln xdx = x (ln x − 1) + C