Question

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

• A. $1004$
• B. $2008$
• C. $2010$
• D. $4020$

Answer 1

The correct option is C $2010$

$I=\int \frac{{\left(\sin x\right)}^{2008}-{\left(\cos x\right)}^{2008}}{{\left(\sin x\right)}^{2008}(\frac{\sin x}{\cos x}+{\left(\frac{\cos x}{\sin x}\right)}^{2009})}dx$

$=\int \frac{\sin x\cos x({\left(\sin x\right)}^{2008}-{\left(\cos x\right)}^{2008})}{{\left(\sin x\right)}^{2010}+{\left(\cos x\right)}^{2010}}dx$

$=\int \frac{({\left(\sin x\right)}^{2009}\cos x-{\left(\cos x\right)}^{2009}\sin x)}{{\left(\sin x\right)}^{2010}+{\left(\cos x\right)}^{2010}}dx$

Put ${\left(\sin x\right)}^{2010}+{\left(\cos x\right)}^{2010}=t$
$\Rightarrow \left(2010\right(\sin x{)}^{2009}\cos x-2010\left(\cos x{)}^{2009}\sin x\right)dx=dt$

$I=\frac{1}{2010}\int \frac{dt}{t}=\frac{1}{2010}\ln |{\left(\sin x\right)}^{2010}+{\left(\cos x\right)}^{2010}|+C$
$\therefore k=2010$