From tangent to cotangent

$What\quad is\quad the\quad value\quad of\quad the\quad integral\quad below:\\ \int _{ \tan { x } }^{ \cot { x } }{ \frac { \ln { { t }^{ 2n } } }{ 1+{ t }^{ 2 } } } dt,\quad where\quad x\in \left( 0,\frac { \pi }{ 2 } \right) \quad and\quad n\quad is\quad a\quad positive\quad integer.\\ \\ Solution:\\ If\quad g:R*\rightarrow R\quad g\left( x \right) =\frac { \ln { { x }^{ 2n } } }{ 1+{ x }^{ 2 } } \quad then\quad \exists G\quad so\quad that\quad G'(x)=g(x)\quad \forall x\in R*\\ Let\quad be\quad f:\left( 0,\frac { \pi }{ 2 } \right) \rightarrow R\quad f(x)=\int _{ \tan { x } }^{ \cot { x } }{ \frac { \ln { { t }^{ 2n } } }{ 1+{ t }^{ 2 } } } dt\\ Then\quad f(x)=G(\cot { x } )-G(\tan { x } )\\ f'(x)=cot'x\cdot g(\cot { x } )-\tan { 'x\cdot } g(\tan { x } )=-(1+\cot ^{ 2 }{ x } )\cdot \frac { \ln { \cot ^{ 2n }{ x } } }{ 1+\cot ^{ 2 }{ x } } -(1+\tan ^{ 2 }{ x } )\cdot \frac { \ln { \tan ^{ 2n }{ x } } }{ 1+\tan ^{ 2 }{ x } } =\\ =-(\ln { \cot ^{ 2n }{ x } } +\ln { \tan ^{ 2n }{ x } } )=-\ln { (\cot ^{ 2n }{ x } } \cdot \tan ^{ 2n }{ x } )=-\ln { 1 } =0\\ So\quad f'(x)=0\quad \forall x\in \left( 0,\frac { \pi }{ 2 } \right) \quad therefore\quad (Lagrange's\quad theorem)\quad f(x)=c\quad \forall x\in \left( 0,\frac { \pi }{ 2 } \right) \quad (c\quad is\quad a\quad constant,\quad a\quad real\quad number)\\ But\quad f(\frac { \pi }{ 4 } )=\int _{ 1 }^{ 1 }{ \frac { \ln { { t }^{ 2n } } }{ 1+{ t }^{ 2 } } } dt=0\\ So\quad finally\quad the\quad constant\quad is\quad 0,\quad so\quad f(x)=0\quad \forall x\in \left( 0,\frac { \pi }{ 2 } \right) \quad and\quad for\quad all\quad n\quad positive\quad integers.\\$

JEE Style..............
Simply put x = pi/4 !!!!

Let us call the integral as $I$

Putting $\large t=\frac{1}{z}$

We get $\large I= \int_{cot(x)}^{tan(x)}\frac{\ln(z^{-2n})}{1+\frac{1}{z^2}}\frac{-1}{z^2} dz$

or $\large I= \int_{tan(x)}^{cot(x)}\frac{\ln(z^{-2n})}{1+z^2}dz$

Which is the same as writing

$\large I= \int_{tan(x)}^{cot(x)}\frac{\ln(t^{-2n})}{1+t^2}dt$

So $\large2I =\int_{tan(x)}^{cot(x)}\frac{\ln(t^{2n}t^{-2n})}{1+t^2}dt$

So $\large2I=\int_{tan(x)}^{cot(x)}\frac{\ln(1)}{1+t^2}dt$

or $\large 2I=0$

Giving $\large I = 0$