$$\int \sec^4 (\theta) d\theta$$
Noting the power of $4$ is even, leave $\sec^2(\theta)$ because this is the derivative of $\tan \theta$ and perform the identity $\tan^2 (\theta)+1=\sec^2 (\theta)$ on the left powers.
$$\sec^4 (\theta)$$
$$(\tan^2 (\theta)+1)\sec^2 (\theta)$$
$$=\tan^2(\theta) \sec^2 (\theta)+\sec^2 (\theta)$$
This can be easily integrated. The first with substituting $u=\tan \theta$ and the second by noting what the derivative of $\tan \theta$ is $\sec^2 (\theta)$
$$\int \tan^2(\theta) \sec^2 (\theta) d\theta+\int \sec^2 (\theta) d\theta$$
$$=\frac{\tan^3 (\theta)}{3}+\tan (\theta)+C$$