Question

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

Answer 1

Answer: 2010

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