.Given. $I=\int_{0}^{\pi / 4}\left(\tan ^{n+1} x\right) d x+\frac{1}{2} \int_{0}^{\pi / 2} \tan ^{n-1}\left(\frac{x}{2}\right) d x$
In second integral, putt $=\frac{x}{2} \Rightarrow d x=2 d t$
$\Rightarrow \quad$ Also, when $x=0$ then $t=0$
When $x=\pi / 2,$ then $t=\pi / 4$
Then, $\begin{aligned} I &=\int_{0}^{\pi / 4}\left(\tan ^{n+1} x\right) d x+\int_{0}^{\pi / 4} \tan ^{n-1} t d t \\ & I=\int_{0}^{\pi / 4} \tan ^{n+1} x \cdot d x+\int_{0}^{\pi / 4} \tan ^{n-1} x d x \\ & \quad\left\{\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(y) d y\right\} \end{aligned}$
$\Rightarrow \quad I=\int_{0}^{\pi / 4}\left(\tan ^{n+1} x+\tan ^{n-1} x\right) d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 4} \tan ^{n-1} x \cdot\left(\tan ^{2} x+1\right) d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 4} \tan ^{n-1} x\left(\sec ^{2} x\right) d x$
Put $\quad t=\tan x$
$\Rightarrow \quad d t=\sec ^{2} x d x$
Also, when $x=0,$ then $t=0$
when $x=\pi / 4,$ then $t=1$ $I=\int_{0}^{1} t^{n-1} d t=\left[\frac{t^{n}}{n}\right]_{0}^{1}=\frac{1}{n}$