Answer
$\frac{3}{4} \tan(4 x) -\frac{5}{4} \sec (4 x) +c.$
Work Step by Step
Since $(\tan 4 x)'=4\sec^2(4x)$ and $(\sec^2(4x))'=4\sec^2(4x)\tan(4x)$, we have
$$
\int \sec 4 x(3 \sec 4 x-5 \tan 4 x) d x=\int3 \sec^2(4 x) -5\sec (4 x)\tan (4 x) d x\\
=\frac{3}{4} \tan(4 x) -\frac{5}{4} \sec (4 x) +c.
$$
