Answer
$$F(t)=\frac{16}{3}t^{3/2}- \sec t +C$$
Work Step by Step
Given $$f(t) = 8\sqrt{t} - \sec t\tan t=8t^{1/2}- \sec t\tan t$$
Then by using table 2 if $f(x)= x^n\ \to\ \ F(x) = \frac{x^{n+1}}{n+1} +C$ and if
$f(x)=\sec x\tan x\ \to\ \ F(x) =\sec x +C$
Hence
\begin{align*}
F(t)&=\frac{8}{3/2}t^{3/2}- \sec t +C \\
&=\frac{16}{3}t^{3/2}- \sec t +C
\end{align*}
To check
\begin{align*}
F'(t) &= 8\sqrt{t} - \sec t\tan t \\
&=f(t)
\end{align*}
