Let $I_n=\int \sec ^n x d x$. If $5 I_6-4 I_4=f(x)$, then $f\left(\frac{\pi}{4}\right)$ is equal to

If

$$ f(x)=\left|\begin{array}{ccc} 1+\sin x+\sin 2 x+\sin 3 x & \frac{3+\sin 2 x}{2} & \frac{-2+\sin 3 x}{3} \\ 3+4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1+\sin x & \frac{1}{2} \sin x & \frac{1}{3} \end{array}\right| $$

then $\int_0^{\pi / 2}\left(f(x)+f^{\prime}(x)\right) d x=$

$$ \lim\limits_{n \rightarrow \infty} \frac{1}{n}\left[\frac{1}{n} \sin ^{-1} \frac{1}{n}+\frac{2}{n} \sin ^{-1} \frac{2}{n}+\ldots+\frac{\pi}{2}\right]= $$

The area (in sq. units) enclosed by the curves $y=2 x-x^2$ and $y=x^2-2 x-6$ is