Answer
$$\frac{1}{5} \tan ^{5} \theta+C$$
Work Step by Step
\begin{aligned} \int \sec ^{2} \theta \tan ^{4} \theta d \theta &=\int\left(\tan ^{2} \theta+1\right) \tan ^{4} \theta d \theta \\ &=\int\left(\tan ^{6} \theta+\tan ^{4} \theta\right) d \theta \\ &=\int \tan ^{6} \theta d \theta+\int \tan ^{4} \theta d \theta \end{aligned}
Use $$\int \tan ^{m} x d x=\frac{\tan ^{m-1} x}{m-1}-\int \tan ^{m-2} x d x$$
Then
\begin{aligned} \int \tan ^{4} \theta d \theta &=\frac{\tan ^{4-1} \theta}{4-1}-\int \tan ^{4-2} \theta d \theta \\ &=\frac{\tan ^{3} \theta}{3}-\int \tan ^{2} \theta d \theta \\ &=\frac{\tan ^{3} \theta}{3}-\left[\frac{\tan ^{2}-1}{2-1}-\int \tan ^{2-2} \theta d \theta\right] \\ &=\frac{\tan ^{3} \theta}{3}-\left[\tan \theta-\int d \theta\right] \\ &=\frac{1}{3} \tan ^{3} \theta-\tan \theta+\theta+C \end{aligned}
and
\begin{aligned} \int \tan ^{6} \theta d \theta &=\frac{\tan ^{6-1} \theta}{6-1}-\int \tan ^{6-2} \theta d \theta \\ &=\frac{\tan ^{5} \theta}{5}-\int \tan ^{4} \theta d \theta \\ &=\frac{\tan ^{5} \theta}{5}-\left[\frac{1}{3} \tan ^{3} \theta-\tan \theta+\theta\right]+C \\ &=\frac{1}{5} \tan ^{5} \theta-\frac{1}{3} \tan ^{3} \theta+\tan \theta-\theta+C \end{aligned}
Hence
\begin{aligned} \int \sec ^{2} \theta \tan ^{4} \theta d \theta=& \frac{1}{5} \tan ^{5} \theta-\frac{1}{3} \tan ^{3} \theta+\tan \theta-\theta \\ &+\frac{1}{3} \tan ^{3} \theta-\tan \theta+\theta+C \\=& \frac{1}{5} \tan ^{5} \theta+C \end{aligned}
