Integration by Parts

Derivation of Integration by Parts

Recall the product rule:

        (uv)' = u'v + uv' 

or  

        uv' = (uv)' - u'v

Integrating both sides, we have that

       

Theorem:  Integration by Parts

Let u and v be differentiable functions, then

         
 



Examples

A)  

u = x du = dx
dv = ex  dx v = ex


Hence we have

       

        =   xex - ex + C

 

Exercise

Evaluate 

       


Integration By Parts Twice

Example

Solve  


       

 
u = x2   du = 2xdx
dv = ex  dx v = ex

 

We have

       

We just did this integral in the last example, so our solution is

        x2ex - 2[xex - ex] + C



The By Parts Trick

Example

Evaluate 

       
u =  ex du = ex dx
dv = sinx dx v = -cosx


We have

       

Now try integration by parts again:


u =  ex du = ex dx
dv = cosx dx v = sinx


       

It seems as if we are back to where we started, however with a clever move, the answer appears.

Let 

       

Then we have

        I = -excosx + ex sinx -  I

Adding I to both sides we get

        2I = -excosx + ex sinx

So that

                    -excosx + ex sinx
        I =                                      +  C
                                2

We can conclude that

       


 

Other By Parts

Example:  

       

u = arctanx                1
du =                   dx
           1 + x2
dv = dx v =  x


We get

       

letting 

        u = 1 + x2,    du = 2x dx,     x dx = du/2

       

Exercise:

Evaluate 

       


 

When to Use Integration By Parts

  1. When there is a mix of two types of functions such as trig and exponentials, poly and trig, etc.

  2. Strange functions such as arctrig and ln.

  3. When all else fails.

 



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