The given function is \(f(\theta ) = \sec \theta \tan \theta \) in the interval \(\left( {0,\frac{\pi }{4}} \right)\).
Obtain the average value of the function, \(f(\theta ) = \sec \theta \tan \theta \) as follows.
The limit values are got from the interval \(\left( {0,\frac{\pi }{4}} \right)\) as \(a = 0\) and \(b = \frac{\pi }{4}\).
Substitute \(a = 0\), \(b = \frac{\pi }{4}\) and \(f(x)\)in the formula \({f_{ave}} = \frac{1}{{b - a}}\int_a^b f (x)dx\) and compute the average value of \(f(x)\) as follows.
\(\begin{aligned}{l}{f_{ave}} &= \frac{1}{{b - a}}\int_a^b f (x)dx\\ &= \frac{1}{{\left( {\frac{\pi }{4} - 0} \right)}}\int_0^{\frac{\pi }{4}} {\sec } \theta \tan \theta d\theta \\ &= \frac{4}{\pi }(\sec \theta )_0^{\frac{\pi }{4}} + C\end{aligned}\)
Further, simplify as follows.
\(\begin{aligned}{l}{f_{{\rm{ave }}}} &= \frac{4}{\pi }\left( {\sec \frac{\pi }{4} - \sec 0} \right)_0^{\frac{\pi }{4}} + C\\ = \frac{4}{\pi }(\sqrt 2 - 1) + C\\ &= \frac{{4(\sqrt 2 - 1)}}{\pi } + C\end{aligned}\)
Thus, the average value of the function \(f(\theta ) = \sec \theta \tan \theta \) in the interval \(\left( {0,\frac{\pi }{4}} \right)\) is \(\frac{{4(\sqrt 2 - 1)}}{\pi } + C\).
