Answer
$$t-2\tan t+c$$
Work Step by Step
Given
$$\int\left(\tan ^{4} t-\sec ^{4} t\right) d t$$
Since
\begin{aligned}
\int\left(\tan ^{4} t-\sec ^{4} t\right) d t&=\int\left(\tan ^{2} t\tan ^{2} t-\sec ^{2} t\sec ^{2} t\right) d t\\
&= \int\left(\tan ^{2} t(\sec ^{2} t-1) -\sec ^{2} t(1+\tan ^{2} t) \right) d t\\
&= \int\left(\tan ^{2} t\sec ^{2} t-\tan^2 t -\sec ^{2} t-\sec ^{2} t\tan ^{2} t \right) d t\\
&= \int\left(\tan ^{2} t\sec ^{2} t-\sec ^{2}+1 -\sec ^{2} t-\sec ^{2} t\tan ^{2} t \right) d t\\
&= \int\left( 1-2\sec ^{2}t \right) d t\\
&=t-2\tan t+c
\end{aligned}
