How do I integrate cosx/(sin^2(x)+9)^2dx?

Question

I'm given the hint: "Make a substitution before using partial fraction decomposition".

edit: squared the denomenator

Answer 1

I partially agree. Yes, to the substitutions, but in order to do PFD, the denominator must be factorable.

It is not.

 

Let u = sinx. Then du = cos x dx.

 

The integral becomes:

 

        du/  (u^2 + 9)

 

Next, 9 is factored out of the denominator:

 

1/9   integeral  du / ( (1/3u)^2 + 1)

 

This is merely the derivative of the arctan function arctan(z) where z = 1/3u

 

Making this substitution z = 1/3u ---> dz = 1/3 du ---> 3 dz = du

 

the integral becomes:

 

1/3 integral dz / (z^2 + 1)

 

1/3 arctan z + k

 

1/3 arctan (1/3u) + k

 

1/3 arctan (1/3 sin (x)) + c