Let $I_n=\int \tan ^n x ~d x, n>1$. If $I_4+I_6=a \tan ^5 x+b x^5+C$, where C is a constant of integration, then the ordered pair $(a, b)$ is equal to

$\left(\frac{1}{5}, 0\right)$

We have, $I_n=\int \tan ^n x d x$

∴  $I_4+I_6=\int \tan ^4 d x+\int \tan ^6 x d x$

$\Rightarrow I_4+I_6=\int\left(\tan ^4 x+\tan ^6 x\right) d x$

$\Rightarrow I_4+I_6=\int \tan ^4 x\left(1+\tan ^2 x\right) d x$

$\Rightarrow I_4+I_6=\int \tan ^4 x \sec ^2 x ~d x=\int \tan ^4 x ~d(\tan x)$

$\Rightarrow I_4+I_6=\frac{1}{5} \tan ^5 x+C$

But, $I_4+I_6=a \tan ^5 x+b x^5+C$

∴  $a=\frac{1}{5}, b=0$

Hence, $(a, b)=\left(\frac{1}{5}, 0\right)$