Evaluate the following integrals:

Let I $=\int \sec ^{4} 2 x d x$

$\Rightarrow I=\int \sec ^{2} 2 x \sec ^{2} 2 x d x$

$\Rightarrow I=\int\left(1+\tan ^{2} 2 x\right) \sec ^{2} 2 x d x$

$\Rightarrow I=\int\left(\sec ^{2} 2 x+\tan ^{2} 2 x \sec ^{2} 2 x\right) d x$

Let $\tan 2 x=t$, then

$\Rightarrow 2 \sec ^{2} 2 x d x=d t$

$\Rightarrow I=\frac{1}{2} \int\left(1+t^{2}\right) d t$

$\Rightarrow I=\frac{1}{2} t+\frac{1}{2} \cdot \frac{1}{3} t^{3}+c$

$\Rightarrow I=\frac{1}{2} \tan 2 x+\frac{1}{6} \tan ^{3} 2 x+c$

Therefore, $\int \sec ^{4} 2 x d x=\frac{1}{2} \tan 2 x+\frac{1}{6} \tan ^{3} 2 x+c$