# −= vdu uv udv

```INTEGRATION BY PARTS
`
udv
=
uv
−
vdu
∫
∫
It is important to see that the above formula is equivalent to the
PRODUCT RULE from Calculus I class (Q: Can you see that?)
PRODUCT RULE:
Some types of problems where you can use the method of integration by parts:
(1) (2) (5) (6) (8) (9) (3) (4) (7) A Collection of Prototypes
prototyprprototypes
TYPE
I: ! &quot;#\$%&amp; ! '( TYPE II: ! &quot;#\$%&amp; ! '( or
! &quot;#\$%&amp; ! '( TYPE III: ! ) &quot;#\$%&amp; '( ! TYPE IV: [ or ]
This uses a special integration by parts method. At the outset, you can
use any one of the following choices:
&amp; '( or &amp; '( Then, at some point, you have to use integration by parts again. They
key is to remember that the second time around, you must use the “same
type of substitution” as the first time. See class notes for the details.
TYPE V: ! *+,&quot;#\$%&amp; *+,'( ! TYPE VI: !- or !- i.e. Odd powers of Sec(x) or Cosec(x)
This uses a special integration by parts method. At the offset, you break
the odd power in to a square and the left over powers, etc. See notes.
```