If I_n=∫limits_0^π / 4 tan ^n θ d θ, then I_8+I_6=

$I_8+I_6=\int\limits_0^{\pi / 4}\left(\tan ^8 \theta+\tan ^6 \theta\right) d \theta$

$=\int\limits_0^{\pi / 4} \tan ^6 \theta \sec ^2 \theta d \theta=\left[\frac{\tan ^7 \theta}{7}\right]_0^{\pi / 4}=\frac{1}{7}$

Hence (4) is the correct answer.