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Techniques of Integration - Change of Variables


Date: 02/17/99 at 17:46:41
From: jenny
Subject: Techniques of integration

Hello.

I would like to know how to solve this question.

   (integral sign) sin 2x/ sq rt (9-cos^4 x)  dx

How should I begin? 

Jen


Date: 02/18/99 at 00:53:14
From: Doctor Luis
Subject: Re: Techniques of integration

You can approach this question more easily if you use the identity 
sin(2x) = 2sin(x)cos(x), and if you rewrite your integral in a more 
suggestive way, like this:

    /
   |     2 sin(x) cos(x)
   | ----------------------- dx
   |  sqrt(9 - (cos^2(x))^2)
  /

At this point it is clear that a change of variables will do the trick.
Let cos^2(x) = 3u. Then, differentiating implicitly,

   2cos(x)*(-sin(x))dx = 3du

This means that 2cos(x)sin(x) = -3du, and so, rewriting our integral in 
terms of u, we have

   /                      /
  |       -3 du           |       du
  |  ---------------- = - | -------------
  |   sqrt(9 - 9 u^2)     |  sqrt(1-u^2)
 /                       /

But this last integral is already known. It's nothing more than the 
inverse cosine function. So,

     /
    |       du
  - | ------------- = arccos(u) = arccos((1/3)cos^2(x))
    |  sqrt(1-u^2)
   /

Notice that we have expressed our final answer in terms of the original 
variable of integration, using the fact that u = (1/3)cos^2(x) .

Obviously, I have neglected the arbitrary constant of integration, but 
you can add that at any time.

Now, to show that the integral of -1/sqrt(1-u^2) is the arccos(u) 
function, what you can do is check that the derivative of the arccos(u) 
function is -1/sqrt(1-u^2). We can do that as follows:

Let y = arccos(x); then x = cos(y). By implicit differentiation on this
last equation you can obtain,

    1 = -sin(y) * dy/dx    (notice we used the chain rule here)

Solving for dy/dx, you get

               -1
    dy/dx = ---------
              sin(y)

Expressing sin(y) in terms of cos(y) (use the identity sin^2(y) + 
cos^2(y) = 1 for this step) we get:

                    -1
   dy/dx  =  -----------------
              sqrt(1-cos^2(y))

But, by definition, x = cos(y). Therefore,

                 -1
   dy/dx  = -------------
             sqrt(1-x^2)

Now, y was just y = arccos(x). Therefore, we have proven that the 
derivative of the arccos(x) function is -1/sqrt(1-x^2) .

Feel free to ask again if you have any other questions.

- Doctor Luis, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Trigonometry

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