First year:
(I)
Integration by parts
Formula and example
It is relatively easy to derive the formula:
u( x)v( x)'  u' ( x)v( x)  u( x)v' ( x)
 u' ( x)v( x)  u( x)v( x)'u( x)v' ( x).
(product rule)
Now, integrate on both sides:
 u ' ( x)v( x)dx   u ( x)v( x)' dx   u ( x)v' ( x)dx
and finally
 u ' ( x)v( x)dx  u ( x)v( x)   u ( x)v' ( x)dx.
The formula above is the formula for integration by parts. With the alternative
notation for the derivative, the formula can be written as follows:

du ( x)
dv( x)
v( x)dx  u ( x)v( x)  
u ( x)dx.
dx
dx
Example:
Calculate
 x sin( x)dx .
The trick consists in choosing wisely what your functions u and v will be.
Here we have to choose u ' ( x)  sin( x) and v( x)  x . Those choices lead to
u ( x)   cos( x) and v' ( x)  1,
so that, applying the formula, we end up with
 x sin( x)dx   x cos( x)   (1 cos( x))dx   x cos( x)  sin( x).
Had we chosen differently, we would have made the problem worse: take
u ' ( x)  x and v( x)  sin( x) this time. This choice leads to
u ( x) 
x2
, v' ( x)  cos( x) 
2
 x sin( x)dx 
x2
1
sin( x)   x 2 cos( x)dx.
2
2
This is mathematically correct, but the integral on the right-hand side is even
more complicated than the one we began with. We have only managed to
make the problem worse.
(II)
Worked examples and exercises
(i)
Calculate  ln( x)dx.
Here, it is not obvious that integration by parts is needed (until the other
possibilities have been exhausted, that is).
We have  ln( x)dx  1 ln( x)dx (it sometimes pays to state the obvious). Let’s
set u ' ( x)  1 and v( x)  ln( x).
We then have u ( x)  x and v ' ( x ) 
1
, so that
x
1
 ln( x)dx  x ln( x)   x  x dx  x ln( x)  x.
(ii)
Calculate
x
5
ln( x) dx .
Similar to the above. Give it a go…
(iii)
Calculate
x
2
sin( x)dx.
Here, set u ' ( x)  sin( x) and v( x)  x 2 . We have u ( x)   cos( x) and v' ( x)  2 x
so that
x
2
sin( x)dx   x 2 cos( x)   2 x cos( x)dx.
It is not a dead end, you just have to use integration by parts again on the
integral on the right-hand side. Finish it…
(iv)
Calculate  cos( x)e x dx.
This example is a classic, but could be quite tricky the first time you do it.
Set u ' ( x)  cos( x) and v( x)  e x . Then u ( x)  sin( x) and v' ( x)  e x , so that
 cos( x)e dx  sin( x)e   sin( x)e dx.
x
x
x
It does not look like much of an improvement, but this time we persevere and
integrate the right-hand side by parts again:
u ' ( x)  sin( x)  u ( x)   cos( x)
, so that
v( x)  e x  v' ( x)  e x
 sin( x)e dx   cos( x)e   cos( x)e dx.
x
x
x
Hence
 cos( x)e dx  sin( x)e
x
x
 cos( x)e x   cos( x)e x dx
so that
 cos( x)e dx  2 sin( x)e
x
1
(v)
Calculate  sin( x)e  x dx.
(vi)
Calculate  2 x cos( x / 2)dx.
x

 cos( x)e x .