That's too much of work

Seriously i didn't know how to solve this, but i used this crazy fact :

Ignore $\log\cot x$ ,because if i am going to use by-parts rule of integration i have to first calculate indefinite integral :

$\displaystyle\int \frac{\sin^{n-1}(2x)}{(\sin^nx + \cos^nx)^2} dx$

which on evaluating i get something of the form $\frac{2^{n-1}}{n^2}g(x,n) + c$

For $n=2009$ it will be $\displaystyle \frac{2^{2008}}{2009^2}g(x,2009) + c$ As per required answer is in the form of $\frac{a^b \log(a)}{c^a}$

On comparing i thought it must be as follows : $a=2 , b=2008 , c=2009$

$a+b+c=4019$

Please Dont laugh..!!! hahahahah!

Open up sin(2theta)

Then divide the expression by (tan(theta) )^2009

Express the integral in the form of log(tan(theta)) and (tan(theta))^2009

Then put (tan(theta))^2009 = t

The expression will be simplified . Integrate the expression using by parts., to get a=2, b=2008, c=2009.