If $I_n =\int\limits_{0}^{π/4}\tan^nθ\, dθ$, then $I_8 +I_6$ equals

$\frac{1}{7}$

We have, $I_n =\int\limits_{0}^{π/4}\tan^nθ\, dθ$

$∴I_8 +I_6=\int\limits_{0}^{π/4}(\tan^8θ+\tan^6θ)dθ$

$⇒I_8 +I_6=\int\limits_{0}^{π/4}\tan^6θ\sec^2θdθ=\int\limits_{0}^{1}t^6\,dt$, where $t=\tan θ$

$⇒I_8 +I_6=\left[\frac{t^7}{7}\right]_{0}^{1}=\frac{1}{7}$