Given: \[ I_n = \int_{0}^{\pi/4} \tan^n \theta \, d\theta \] Rewrite \( I_n \) as: \[ I_n = \int_{0}^{\pi/4} \tan^{n-2} \theta \cdot \tan^2 \theta \, d\theta \] Using \(\tan^2 \theta = \sec^2 \theta - 1\): \[ I_n = \int_{0}^{\pi/4} \tan^{n-2} \theta \cdot (\sec^2 \theta - 1) \, d\theta \] Expanding: \[ I_n = \int_{0}^{\pi/4} \tan^{n-2} \theta \cdot \sec^2 \theta \, d\theta - \int_{0}^{\pi/4} \tan^{n-2} \theta \, d\theta \] Substituting \( \tan \theta = t \), so \( \sec^2 \theta \, d\theta = dt \), the limits become \( \tan 0 = 0 \) and \( \tan(\pi/4) = 1 \): \[ I_n + I_{n-2} = \int_{0}^{1} t^{n-2} \, dt \] Solve the integral: \[ \int_{0}^{1} t^{n-2} \, dt = \left[\frac{t^{n-1}}{n-1}\right]_{0}^{1} = \frac{1}{n-1} \] Thus: \[ I_n + I_{n-2} = \frac{1}{n-1} \] For \( n = 8 \): \[ I_8 + I_6 = \frac{1}{7} \]