Class Notes (1,100,000)
CA (630,000)
UTSG (50,000)
MAT (4,000)
MAT136H1 (900)
all (200)
Lecture

MAT136H1 Lecture Notes - Product Rule

Department
Mathematics
Course Code
MAT136H1
Professor
all

Page:
of 1 7.1 Integration Techniques
Integration by Parts
Question #1 (Easy): Solving Indefinite Integral
Strategy
Integration by parts is useful when the integral contains two seemingly different function that cannot
further be simplified, so that:   .
This is after the form of the Product Rule. Assign simpler variables like    and let   ,
then by taking derivative of ,    and by taking the anti-derivative of ,  .
Sample Question
Evaluate the integral using integration by parts.
 

Solution
More complex function that simplifies when its derivative is taken should be assigned to . So here,
   . Then   
. Taking derivative of gives   , and taking the anti-
derivative of  gives  

 

.
Given the form of integration by parts:
     
Here:  
  

 

  . Since the second integral
has two functions that cannot be simplified, integration by parts needs to be taken one more time.
So again let   , and 

. Then,    and   
 
.
Then simplify the second integral:

  
 
 
 
  

 

  

 

  

 

  .
Merge this into the first part of the answer:  

  

 

 
 
  

 

  
  

 

  
Therefore, by integration by parts the integral is evaluated as:
 

  

 

  