If \(I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x\), where \(n\) is a positive integer, then \(I_{10}+I_{8}\) is

Given integral is,
\[
\begin{aligned}
I_{n} &=\int_{0}^{\pi / 4} \tan ^{n} x d x \\
I_{n} &=\int_{0}^{\pi / 4} \tan ^{n-2} x \cdot \tan ^{2} x d x \\
&=\int_{0}^{\pi / 4} \tan ^{n-2} x \cdot\left(\sec ^{2} x-1\right) d x \\
I_{n} &=\int_{0}^{\pi / 4} \tan ^{n-2} x \cdot \sec ^{2} x d x \\
&-\int_{0}^{\pi / 4} \tan ^{n-2} x d x
\end{aligned}
\]
\[
\begin{gathered}
I_{n}=\int_{0}^{1} t^{n-2} d t-\int_{0}^{\pi / 4} \tan ^{n-2} x d x \\
I_{n}=\left[\frac{t^{n-1}}{n-1}\right]_{0}^{1}-I_{n-2}\left\{\begin{array}{l}
\text { put } \tan x=t \\
\Rightarrow \sec ^{2} x d x=d t
\end{array}\right\} \\
I_{n}+I_{n-2}=\frac{1}{n-1}
\end{gathered}
\]
Put \(n=10\), we get
\[
I_{10}+I_{8}=\frac{1}{9}
\]