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Integration by Parts

Date: 04/08/99 at 00:13:10
From: Paul Auge
Subject: Integration by parts

Trying to solve the following integral, I wind up in a continuous loop 
or with everything equal to zero.

INT[(sec x)^3 dx]

Please help.


Date: 04/08/99 at 15:27:27
From: Doctor Bruce
Subject: Re: Integration by parts

Hello Paul,

Integration by parts will simplify this problem a little. Set it up 
this way:

      u = sec x                dv = (sec x)^2 dx

     du = (sec x)(tan x) dx     v = tan x

Then you will get

               INT[u dv]  =      uv        - INT [v du]

 INT[(sec x)^3 dx] = (sec x)(tan x) - INT[(sec x)(tan x)^2 dx].

The trick is now to use the trig identity tan^2 = sec^2 - 1 under the
integral on the right side. After rearranging things a little you get

     2*INT[(sec x)^3 dx] = (sec x)(tan x) + INT[(sec x) dx].

So now we have to figure out the integral of secant, rather than the
integral of secant-cubed.

Unfortunately, I don't know any easy way to explain how to evaluate 
the integral of secant.  I'll just have to give you the answer:

     INT[(sec x) dx] = log[(1 + sin x)/cos x].

That's not very satisfying, I know. It always annoyed me when the 
teacher pulled a magic formula out of thin air. But there it is. You 
can take the derivative of the right side and see that you do get 
sec x, so that checks.

Integration by parts would not help in integrating secant, since the 
answer involves logs. Remember that integration by parts only involves 
products. Of course, you don't know what the answer is supposed to 
look like! So you try integration by parts and end up chasing your 
tail sometimes, as you discovered.

I hope this helps.

- Doctor Bruce, The Math Forum   

Date: 04/08/99 at 15:31:18
From: Doctor Rob
Subject: Re: Integration by parts

Thanks for writing to Ask Dr. Math!

This seemingly simple integral is surprisingly hard. The proper
substitution is

   z = tan(x/2),
   x = 2*arctan(z),
   dx = 2*dz/(1+z^2),
   sec(x) = (1+z^2)/(1-z^2).

That transforms the integral into

   INT[2*(1+z^2)^2/(1-z^2)^3 dz].

This you can split using partial fractions into

   (1/2)*INT[1/(1-z) - 1/(1-z)^2 + 2/(1-z)^3 + 1/(1+z) - 1/(1+z)^2 +
             2/(1+z)^3 dz].

Each of these terms is easily integrated. When you are done, you have 
to resubstitute z = tan(x/2) and simplify.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Calculus

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