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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

mathlove
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DXT
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  • Could you please tell me source of this problem, thank – King Tut Jan 18 '18 at 15:24
  • Might be easier if you put n instead of 2009. – marty cohen Feb 02 '18 at 16:11
  • Actually original question was Finding $\displaystyle \int\frac{\ln(\cot x)}{\bigg(\sin^{2009}x+\cos^{2009} x\bigg)^2}\cdot (\sin^{2008}(2x))dx$extremelly sorry for that. – DXT Feb 03 '18 at 03:25
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    Do you wish to change original question or keep it in original form? Math love has edited and even answered with edited question. Edited question can be said to be much easier. – King Tut Feb 03 '18 at 06:08

1 Answers1

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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 $$\int\frac{\ln(\cot x)}{(\sin^{2009}x+\cos^{2009} x)^2}\cdot (\sin^{2008}(2x))\ \mathrm dx$$$$=\frac{2^{2008}}{2009^2}\left(\frac{\ln(\cot^{2009}x)}{1+\cot^{2009}x}+\ln(1+\cot^{2009}x)-\ln(\cot^{2009}x)\right)+\mathrm C$$

mathlove
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