Given that,
$I_n=\int \sec ^n x d x$
$f(x)=5 I_6-4 I_4=5 \int \sec ^6 x d x-4 \int \sec ^4 x d x$
$=\int 5 \sec ^4 x \sec ^2 x d x-\int 4 \sec ^2 x \sec ^2 x d x$
$f(x)=\int\left\{5\left(1+\tan ^2 x\right)^2-4\left(1+\tan ^2 x\right)\right\} \sec ^2 x d x$
Let $u=\tan x$
Differentiating w.r.t. ' $x$ ', we get
$d u=\sec ^2 x d x$
$\therefore \quad f(x)=\int\left\{5\left(1+u^2\right)^2-4\left(1+u^2\right)\right\} d u$
$=\int\left\{5\left(1+u^4+2 u^2\right)-4\left(1+u^2\right)\right\} d u$
$=\int\left(5+5 u^4+10 u^2-4-4 u^2\right) d u$
$=\int\left(5 u^4+6 u^2+1\right) d u=5 \frac{u^5}{5}+6 \frac{u^3}{3}+u$
$f(x)=u^5+2 u^3+u$
$f(x)=(\tan x)^5+2(\tan x)^3+\tan x$
At $x=\frac{\pi}{4}, f\left(\frac{\pi}{4}\right)=\left(\tan \frac{\pi}{4}\right)^5+2\left(\tan \frac{\pi}{4}\right)^3+\tan \frac{\pi}{4}$
$=(1)^5+2(1)^3+(1)$
$f\left(\frac{\pi}{4}\right)=4$