Integration by Parts

1. Derivation of Integration by Parts
   Recall the product rule:

   (uv)' = u'v + uv' or  uv' = (uv)' - u'v

   Integrating both sides, we have that

   int uv' dx = int (uv)' dx- int u'v dx

   = uv - int u'v dx.

   Theorem:  Integration by Parts

   Let u and v be differentiable functions, then
   

   
   

2. Examples

   A)  
   

   u = x	du = dx
   dv = ex  dx	v = ex

   Hence we have

   

   =  xe^x -e^x + C

   Exercise:   Evaluate int xsinx dx
   

3. Integration By Parts Twice

   Example:  

   Solve  int x² e^x dx

    

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

   We have

   x²e^x - int 2xe^x dx = x²e^x - 2int xe^x dx

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

   x²e^x - 2[xe^x -e^x] + C
   

4. The By Parts Trick

   Example:

   Evaluate int e^xsinx dx
   

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

   We have

   int e^xsinx dx = -e^xcosx + int e^xcosx dx
   

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

   int e^xsinx dx =  -e^xcosx + e^x sinx - int e^xsinx dx

   Let I = int e^xsinx dx

   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

   I = (-e^xcosx + e^x sinx)/2 + C

   We can conclude that

   int e^xsinx dx = (-e^xcosx + e^x sinx)/2 + C
   

5. Other By Parts

   Example:  int arctan x dx
   

   u = arctanx	du = 1/(1 + x2) dx
   dv = dx	v =  x

   We get

   xarctanx - int x/((1 + x²) dx

   letting u = 1 + x², du = 2xdx, xdx = du/2

   xarctanx - 1/2 int 1/u du  =  xarctanx - 1/2 ln|(1 + x²)| + C

   Exercise:

   Evaluate int ln x dx
   

6. 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.