Answer
\[1-\frac{2}{\pi}\ln 2\]
Work Step by Step
Let \[I=\int x\sec^2 x\;dx\]
Using integration by parts:
\[I=x\in \sec^2 x\;dx-\int\left[\frac{d}{dx}(x)\int \sec^2 xdx\right]dx\]
\[I=x\tan x-\int (1)\tan x\;dx\]
\[I=x\tan x-\int \tan x\;dx\]
\[I=x\tan x+\ln |\cos x|\]
\[\Rightarrow \int_{0}^{\frac{\pi}{4}} x\sec^2 x\;dx=\left[I\right]_{0}^{\frac{\pi}{4}}\]
\[\Rightarrow \int_{0}^{\frac{\pi}{4}} x\sec^2 x\;dx=\left[x\tan x+\ln |\cos x|\right]_{0}^{\frac{\pi}{4}}\]
\[\Rightarrow \int_{0}^{\frac{\pi}{4}} x\sec^2 x\;dx=\left[\frac{\pi}{4}(1)+\ln\left|\frac{1}{\sqrt 2}\right|\right]-[0+\ln 1]\]
\[\Rightarrow \int_{0}^{\frac{\pi}{4}} x\sec^2 x\;dx=\frac{\pi}{4}-\frac{1}{2}\ln 2\]
Average value of function $f(x)$ on the intevral $[a,b]$ is given by:
\[f_{average}=\frac{1}{b-a}\int_a^b f(x)dx\]
Therefore the required average value is given by:
\[f_{average}=\frac{1}{\frac{\pi}{4}-0}\int_0^{\frac{\pi}{4}} x\sec^2 xdx\]
\[\Rightarrow f_{average}=\frac{4}{\pi}\left[\frac{\pi}{4}-\frac{1}{2}\ln 2\right]=1-\frac{2}{\pi}\ln 2\]
