Answer
$$\frac{1}{3} \tan ^{3} \theta+ \tan\theta+C $$
Work Step by Step
\begin{aligned} \int \frac{d \theta}{\cos ^{4} \theta} &=\int \sec ^{4} \theta d \theta \\ &=\int \sec ^{2} \theta \cdot \sec ^{2} \theta d \theta \\ &=\int\left(\tan ^{2} \theta+1\right) \cdot \sec ^{2} \theta d \theta \\
&= \int \tan ^{2} \theta \sec ^{2} \theta d \theta+\int \sec ^{2} \theta d \theta\\
&=\frac{1}{3} \tan ^{3} \theta+ \tan\theta+C
\end{aligned}
