Solution 5.

See text and/or instructor's solution manual.

Answer.   .  

Solution.  The integrand is  ,  construct the complex function

using the substitutions    and  .

The complex integrand is  ,  and we have  

                      

The complex integrand is  ,  and we have  

                    .  

Factoring the denominator, we obtain

                      

Hence  ,  has a zero of order  2  at    and simple zeros at     and     lie inside  .

Hence  ,  has a pole of order  2  at    and simple poles at     and     lie inside  .

                    The points    lie inside  .

Using Theorem 8.1 (Cauchy's Residue Theorem), the value of the integral is computed   

                      

Here the denominator of  f(z) has a factor of the form  ,  and by Theorem 8.2  .  

In this exercise, the limit can be calculated as follows:

                        

Here the denominator of  f(z) has a factor of the form  ,  and by Theorem 8.2  .  

In this exercise, the limit can be calculated as follows:

                        

We are done.   

Aside.  We can let Mathematica double check our work.

And we can let Maple check our work too.

     > V1 := residue( I*(z^2-1)^2/(4*z^2*(2*z^2+5*z+2)), z=0 );

               

     > V2 := residue( I*(z^2-1)^2/(4*z^2*(2*z^2+5*z+2)), z=-1/2 );

               

     > 2*Pi*I*(V1+V2);

               

          The area under the curve    is  .  

We are really done.

Aside.  Both and are capable of finding the definite integral.

          > int((sin(t))^2/(5+4*cos(t)),t=0..2*Pi);  

                              

We are really really done.

Caveat.  Both and can find functions for the indefinite integral.  These functions are discontinuous

and will produce erroneous computations for the value of the definite integral when using the Fundamental Theorem of Calculus.

Different versions of Mathematica will find the following indefinite integrals:  

Maple will find the following indefinite integral:  

          > int((sin(t))^2/(5+4*cos(t)),t);  

                              

The failed attempts to find the definite integral using   are summarized:

Warning.  The above answers are wrong because the indefinite integrals are discontinuous.

The indefinite integral    is discontinuous at the endpoints  .
Thus the calculation    is not defined.  
However     is a valid calculation.  

The indefinite integral    is discontinuous at the point  .
Thus the calculation    is a wrong answer.  
However     is a valid calculation.  

The indefinite integral    is discontinuous at the point  .
Thus the calculation    is a wrong answer.  
However     is a valid calculation.  

Question.  The definite integral subroutines in Mathematica and Maple were able to compute the definite integral correctly,

but using the formula    for the indefinite integral required the additional use of one sided limits.

How do you think the software was able to detect this situation and come up with the proper answer?  

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell