1. We develop the formula by considering how to differentiate products.
where u and v are both functions of x.
Substituting for y,
e.g. If ,

2. So,
Integrating this equation, we get
The l.h.s. is just the integral of a derivative, so, since integration is the reverse of differentiation, we get
Can you see what this has to do with integrating a product?

3. The function in the integral on the l.h.s. . . .
Here’s the product . . .
if we rearrange, we get
The function in the integral on the l.h.s
. . . is a product, but the one on the r.h.s
is a simple function that we can integrate easily.

4. Here’s the product . . .
if we rearrange, we get
So, we’ve integrated !
We need to turn this method ( called integration by parts ) into a formula.

5. Example
Generalisation
Integrating:
Simplifying the l.h.s.:
Rearranging:

6. SUMMARY
Integration by Parts
To integrate some products we can use the formula

7. So,
Using this formula means that we differentiate one factor, u
to get

8. So,
Using this formula means that we differentiate one factor, u
to get
and integrate the other ,
to get v

9. So,
Using this formula means that we differentiate one factor, u
to get
and integrate the other ,
to get v
e.g. 1 Find
Having substituted in the formula, notice that the 1st term, uv, is completed but the 2nd term still needs to be integrated.
and
differentiate
integrate
( +C comes later )

10. So,
differentiate
integrate
and
We can now substitute into the formula

11. So,
and
differentiate
integrate
We can now substitute into the formula
The 2nd term needs integrating

12. e.g. 2 Find
Solution:
and
differentiate
This is a compound function, so we must be careful.
integrate
So,

13. Exercises
Find 1. 2. 1.
Solutions: = ò dx xe x 2.

14. Definite Integration by Parts
With a definite integral it’s often easier to do the indefinite integral and insert the limits at the end.
We’ll use the question in the exercise you have just done to illustrate.

15. Using Integration by Parts
Integration by parts cannot be used for every product.
It works if
we can integrate one factor of the product,
the integral on the r.h.s. is easier* than the one we started with.
* There is an exception but you need to learn the general rule.

16. e.g. 3 Find
Solution:
What’s a possible problem?
ANS: We can’t integrate
Can you see what to do?
If we let and , we will need to differentiate and integrate x.
Tip: Whenever appears in an integration by parts we choose to let it equal u.

17. x cancels. e.g. 3 Find So, integrate differentiate
The r.h.s. integral still seems to be a product!
BUT . . .
x cancels.
So,

18. e.g. 4
Solution:
Let
and
The integral on the r.h.s. is still a product but using the method again will give us a simple function.
We write

19. e.g. 4
Solution:
( 1 )
Let
and
So,
Substitute in
( 1 )

20. Example
e.g. 5 Find
Solution:
It doesn’t look as though integration by parts will help since neither function in the product gets easier when we differentiate it.
However, there’s something special about the 2 functions that means the method does work.

21. e.g. 5 Find
Solution:
We write this as:

22. e.g. 5 Find
So,
where
and
We next use integration by parts for I2

23. e.g. 5 Find
So,
where
and
We next use integration by parts for I2

24. 2 equations, 2 unknowns ( I1 and I2 ) !
e.g. 5 Find
So,
( 1 )
( 2 )
2 equations, 2 unknowns ( I1 and I2 ) !
Substituting for I2 in ( 1 )

25. 2 equations, 2 unknowns ( I1 and I2 ) !
e.g. 5 Find
So,
( 1 )
( 2 )
2 equations, 2 unknowns ( I1 and I2 ) !
Substituting for I2 in ( 1 )

26. 2 equations, 2 unknowns ( I1 and I2 ) !
e.g. 5 Find
So,
( 1 )
( 2 )
2 equations, 2 unknowns ( I1 and I2 ) !
Substituting for I2 in ( 1 )

27. Exercises 1. 2.
( Hint: Although 2. is not a product it can be turned into one by writing the function as )

28. Solutions:
and
Let 1.
( 1 )
and
Let
For I2:
Subs. in ( 1 )

29. 2.
This is an important application of integration by parts
and
Let
So,