Possible title: How do I solve this definite integral evaluation?

[juantheron]

Evaluation of \(\displaystyle \displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\ln(\cot x)}{\left[(\sin x)^{2009}+(\cos x)^{2009}\right]^2}\cdot (\sin 2x)^{2008}dx\)

What I have Tried:: Let \(\displaystyle \displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\ln(\cot x)}{\left[(\sin x)^{2009}+(\cos x)^{2009}\right]^2}\cdot (\sin 2x)^{2008}dx\)

So \(\displaystyle \displaystyle I = \int_{0}^{\frac{\pi}{4}}\frac{\ln(\cot x)}{(\cos x)^{4018}\left[1+(\tan x)^{2009}\right]}\cdot 2^{2008}\cdot (\sin x)^{2008}\cdot (\cos x)^{208}dx\)

So $\displaystyle I = 2^{2008}\int_{0}^{\frac{\pi}{4}}\frac{\ln(\cot x)\cdot (\tan x)^{2008}\cdot \sec^2 x}{\left[1+(\tan x)^{2009}\right]^2}dx\;,$

Now put $\tan x= t$ and $\sec^2 dx = dt$ and changing Limits , We get

$\displaystyle I = -2^{2008}\int_{0}^{1}\frac{\ln (t)\cdot t^{2008}}{[1+t^{2009}]^2}dt$

Now How can I solve after that, Help me

Thanks

 

[alyafey22]

The numerical value of the integral

$$\int_{0}^{1}\frac{\ln (t)\cdot t^{2008}}{[1+t^{2009}]^2}dt$$

using wolframalpha is very small 1.71738×10^-7. Are you sure about the limits of the integration ?