Question

If the value $\int \frac{1-(\cot x)^{2008}}{\tan x+(\cot x)^{2009}} d x=\frac{1}{k} \ln \left|\sin ^k x+\cos ^k x\right|+C$, then find $k$.

Answer 1

L.H.S. $\int \frac{(\sin x)^{2008}-(\cos x)^{2008}}{(\sin x)^{2008}\left(\frac{\sin x}{\cos x}+\left(\frac{\cos x}{\sin x}\right)^{2009}\right)} d x$
$=\int \frac{\sin x \cos x\left((\sin x)^{2008}-(\cos x)^{2008}\right)}{(\sin x)^{2010}+(\cos x)^{2010}} d x $
$=\int \frac{\left((\sin x)^{2009} \cos x-(\cos x)^{2009} \sin x\right)}{(\sin x)^{2010}+(\cos x)^{2010}} d x $
$\text { put }(\sin x)^{2010}+(\cos x)^{2010}= t$
$=\frac{1}{2010} \int \frac{ dt }{ t } $
$=\frac{1}{2010} \ln \left|(\sin x)^{2010}+(\cos x)^{2010}\right|+c $
$\Rightarrow k =2010 $