AP BC Calc
–
4.1 (cont’d cont’d)
Integration by Parts and Trigonometric Integrals
Ferguson Notes
In the first part of 4.1, the part addressing the method of Integration by Substitution, we found that the Substitution
Rule was a response to differentiation by the Chain rule. That is, to integrate something that was differentiated by the
Chain Rule, we
must use Integration by Substitution.
But what do we do if we’re faced with something that was
differentiated using the Product (or Quotient) Rule?
To integrate products (which includes quotients), we use a method called
Integration by Parts
.
Note that the product rule says the following:
( ) ( )
( ) ( )
( )
( )
f x g x
f
x g x
f x g
x
.
Therefore,
( ) ( )
( )
( )
( ) ( )
f
x g x
f x g
x
dx
f x g x
C
.
This equation can be rearranged as such:
( )
( )
( ) ( )
( ) ( )
f x g
x dx
f x g x
f
x g x dx
.
Is that much better?
No?
Fine, then let
( )
f x
u
and
( )
g x
v
, which means
( )
du
f
x dx
and
( )
dv
g x dx
. By substitution, the
equation becomes much easier to see and understand (and remember):
udv
uv
vdu
Integration by parts is pretty much that….substitution.
The hardest part, in my point of view is choosing the right
functions for u and dv…..
Ex. 1
–
find
sin
x
xdx
.
At this point, we can see that using the Substitution Rule is going to do us no good in evaluating this integral, so the next
best method to try in integration by parts. We also knew that we’d need to integrate this by parts because it is a product
of two functions (not always a reliable method of remembering, though).
If we let
u
x
and
sin
dv
x
, then
du
dx
and
cos
v
x
.
So by utilizing our integration by parts method,
sin
cos
cos
cos
cos
cos
sin
x
xdx
x
x
xdx
x
x
xdx
x
x
x
C
.
Done.
“Well that was easy.”
Ex. 2
–
find
ln
xdx
.
Let
ln
u
x
and
dv
dx
, then
1
du
x
and
v
x
.
1
ln
ln
ln
ln
xdx
x
x
x
dx
x
x
dx
x
x
x
C
x
.
Wow
–
these integrals are going quickly!