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An Integration with Substitution


Date: 06/06/98 at 22:05:28
From: Abecina
Subject: Integrations 

Hello Dr Math, 

I am having a few problems with some integrations. Will you please 
help me?

1) Use the result of INT(0 to a) f(x) dx = INT (0 to a) f(a-x) dx to
   show that:

                  x sin(x)        pi^2
   INT(0 to pi) ------------ dx = ----
                1 + cos^2(x)       4

2) If:

                        x^n
   I(n) = INT(0 to 1) ------- dx
                      x^2 + 1

                                  1
   i) Show that I(n) + I(n-2) = -----
                                n - 1

   ii) Evaluate I(4) and I(5).

Thank you.


Date: 06/07/98 at 07:53:39
From: Doctor Anthony
Subject: Re: Integrations 

Problem 1:

Since sin(pi-x) = sin(x)  and cos(pi-x) = -cos(x), we can use the 
above theorem to write the following equation:

           x sin(x) dx        (pi-x) sin(x) dx
   I = INT ------------ = INT ----------------
           1 + cos^2(x)         1 + cos^2(x)

Then adding these two values of I we get:

            pi sin(x)dx
   2I = INT ------------
            1 + cos^2(x)

Let cos(x) = u. Thus -sin(x) dx = du, and x = 0 implies u = 1 and 
x = pi implies u = -1. So:

                      -pi du                  pi du
   2I =  INT(1 to -1) ------- = INT(-1 to 1) -------
                      1 + u^2                1 + u^2

   2I = pi tan^(-1)(u) from -1 to 1

      = pi [pi/4 - (-pi/4)]

      = pi [pi/2]

and so:

   I = pi^2/4  as required.

Problem 2:
                                  
(i) We have:

                         x^n + x^(n-2)
   I(n) + I(n-2) =  INT [-------------]dx
                            x^2 + 1  

                        x(n-2)(x^2 + 1)
                 = INT[----------------]dx
                            x^2 + 1

                 = INT[x^(n-2) dx]   from 0 to 1

                   x^(n-1)
                 = -------    from 0 to 1
                    n - 1

                     1
                 = -----     as required
                   n - 1                    

(ii) Evaluate I(4) and I(5).

   I(4) + I(2) = 1/3

   I(2) + I(0) = 1/1

and 

   I(0) = INT[dx/(1+x^2)]  = tan^(-1)(x)  from 0 to 1

        =  pi/4

   I(2) = 1 - pi/4

   I(4) = 1/3 - (1 - pi/4)

   I(4) = pi/4 - 1 + 1/3  =  pi/4 - 2/3                    

For I(5):

   I(5) + I(3) = 1/4

   I(3) + I(1) = 1/2

                 
and 
               x dx
   I(1) = INT[-------]
              1 + x^2       

Let 1+x^2 = u. Thus 2x dx = du and x dx = (1/2) du. So:

   I(1) = (1/2)INT[du/u]  where x=0 implies u=1 and x=1 implies u=2

        = (1/2) ln(u)  = (1/2)[ln(2)- ln(1)]

        = (1/2) ln(2)

So I(5) = 1/4 - [1/2 - (1/2)ln(2)]

        = (1/2)ln(2) - 1/2 + 1/4

        = (1/2)ln(2) - 1/4      

-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus

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