Trig Identities Simplify a Challenging Integration

Date: 02/11/2005 at 03:44:27
From: Jason
Subject: A hard trig integral!

Hi,

How would you integrate: 1/[asin(theta) + bcos(theta)]^2 ?

I tried to multiply the denominator out and got down to:

  (a^2)(sintheta)^2 + (b^2)(costheta)^2 + ab sin(2theta)

I have no idea where to go next!  It's tempting when you see the 
(sintheta)^2 + (costheta)^2 = 1, but in this case, the constants in 
front of them are not equal!  Is there any way out of this?

Date: 02/11/2005 at 13:09:33
From: Doctor Ash
Subject: Re: A hard trig integral!

Hi Jason,

I must admit that this is a really interesting integral.  This one
might look simple, but we somehow get stuck when we try to solve it.
The following explanation will help analyze the integral and then
solve it.  For simpler readability, I will use x instead of theta.

First of all, it seems that there is nothing we can do but to multiply
out the denominator, as you did.  So we have in the denonminator:

  a^2(sin x)^2 + 2ab(sin x)(cos x) + b^2(cos x)^2

The question is that what can we do next?  You are right when you said
that the Pythagorean identity is tempting, but will not work here.  
One thing we can do is to factor.  If we factor out b^2(cos x)^2, we 
will end up with:

  b^2(cos x)^2[(a^2/b^2)(tan x)^2 + (2a/b)(tan x) + 1]

This follows from the trigonometric identity:
  
  (sin x)/(cos x) = tan x

This might look even more complicated, and you might wonder how this
would help us.  In fact, it does not help us unless we recognize that
the second factor is an expansion of:

  [(a/b)(tan x) + 1]^2

Now that we can deal with!  But before we do, there is one more
conversion we can do.  Recall that:

  1/(cos x)^2 = (sec x)^2

So the final form of the integral is:

  int [(sec x)^2 /b^2((a/b)tan x + 1)] dx

Now we use the u-substitution.

  Let   u = (a/b)tan x + 1
       du = (a/b)(sec x)^2 dx
  (b/a)du = (sec x)^2 dx

The integral simplifies to:

   int [(b/a)/(b^2)(u^2)] du       Note: a and b are constants

Now, this is an integral we can do.  Do not forget to substitute for 
u at the end, and remember to include the constant of integration C 
since this is an indefintie integral.

This is how this integral can be solved.  I hope this explanation was 
helpful and if you have any questions, please feel free to write me back.


- Doctor Ash, The Math Forum
  http://mathforum.org/dr.math/