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

Examples

A)  

u = x	du = dx
dv = ex  dx	v = ex

Hence we have

        =   xe^x - e^x + 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

        x²e^x - 2[xe^x - e^x] + 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 = -e^xcosx + e^x sinx -  I

Adding I to both sides we get

        2I = -e^xcosx + e^x sinx

So that

                    -e^xcosx + e^x 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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