Finding $\displaystyle \int\frac{\ln(\cot x)}{\bigg(\sin^{2009}x+\cos^{2009} x\bigg)^2}\cdot (\sin^{2008}(2x))dx$
Try: $$\int \frac{\ln(\cot x)}{\bigg(1+\tan^{2009}(x)\bigg)^2}\cdot \tan^{2008}(x)\cdot \sec^{2010}(x)dx$$
Now substuting $\tan^{2009} x=t$ and $\displaystyle \tan^{2008}(x)\cdot \sec^2(x)dx=\frac{1}{2009}dt$
So $$ -\frac{1}{(2009)^2}\int\frac{\ln(t)}{(1+t)^2}\cdot (1+t^{\frac{2}{2009}})^{1005}dt$$
Could some help me to solve it, thanks
Using $\sin(2x)=2\sin x\cos x$, we have $$\begin{align}&\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot (\sin^{n}(2x))\ \mathrm dx\\\\&=\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot 2^n\sin^nx\cos^nx\ \mathrm dx\\\\&=-2^n\int\frac{\ln(\cot x)}{(\sin^{n+1}x+\cos^{n+1} x)^2}\cdot \sin^{2n+2}x\cdot\frac{-\cot^nx}{\sin^2x}\ \mathrm dx\\\\&=-2^n\int\frac{\ln(\cot x)}{(1+\cot^{n+1} x)^2}\cdot\frac{-\cot^nx}{\sin^2x}\ \mathrm dx\end{align}$$
Let $s=1+\cot^{n+1}x$. Then, since $\frac{\mathrm ds}{n+1}=\frac{-\cot^nx}{\sin^2x}\ \mathrm dx$, we get
$$\begin{align}&-2^n\int\frac{\ln(\cot x)}{(1+\cot^{n+1} x)^2}\cdot\frac{-\cot^nx}{\sin^2x}\ \mathrm dx\\\\&=-2^n\int\frac{\frac{1}{n+1}\ln(s-1)}{s^2}\cdot\frac{\mathrm ds}{n+1}\\\\&=\frac{2^n}{(n+1)^2}\int (s^{-1})'\ln(s-1)\ \mathrm ds\\\\&=\frac{2^n}{(n+1)^2}\left(s^{-1}\ln(s-1)-\int\frac{1}{s(s-1)}\ \mathrm ds\right)\\\\&=\frac{2^n}{(n+1)^2}\left(s^{-1}\ln(s-1)+\int\left(\frac{1}{s}-\frac{1}{s-1}\right)\ \mathrm ds\right)\\\\&=\frac{2^n}{(n+1)^2}\left(s^{-1}\ln(s-1)+\ln s-\ln(s-1)\right)+\mathrm C\\\\&=\frac{2^n}{(n+1)^2}\left(\frac{\ln(\cot^{n+1}x)}{1+\cot^{n+1}x}+\ln(1+\cot^{n+1}x)-\ln(\cot^{n+1}x)\right)+\mathrm C\end{align}$$
Therefore, for $n=2008$, we have $$\int\frac{\ln(\cot x)}{(\sin^{2009}x+\cos^{2009} x)^2}\cdot (\sin^{2008}(2x))\ \mathrm dx$$$$=\frac{2^{2008}}{2009^2}\left(\frac{\ln(\cot^{2009}x)}{1+\cot^{2009}x}+\ln(1+\cot^{2009}x)-\ln(\cot^{2009}x)\right)+\mathrm C$$