Trigonometric Integration! (7)

$\large \displaystyle I_n = \int_0^{\pi/4} \tan ^n x .dx \implies I_{n+2} = \int_0^{\pi/4} \tan ^{n+2} x.dx\\ \large \displaystyle \therefore I_n + I_{n+2} = \int_0^{\pi/4} (\tan ^n x + \tan ^{n+2} x). dx\\ \large \displaystyle = \int_0^{\pi/4} \tan ^n x (1+\tan^2 x) . dx\\ \large \displaystyle = \int_0^{\pi/4} \tan ^n x \times \sec^2 x . dx\\ \large \displaystyle \text{If } t = \tan x, dt = \sec ^2 x.dx\\ \large \displaystyle \text{When } x = 0, t = 0 \text{ and } x = \frac{\pi}{4}, t = \tan \frac{\pi}{4} = 1\\ \large \displaystyle \implies I_n + I_{n+2} = \int_0^1 t^n . dt = \left[ \frac{t^n + 1}{n+1} \right]_0^1\\ \large \displaystyle \therefore I_n + I_{n+2} = \color{#D61F06}{\boxed{\frac{1}{n+1}}}.$