## Calculus (3rd Edition)

$\frac{1}{4} (\tan 2\theta)^2+c$
Since $u=\tan 2\theta$, then $du=2\sec^2 2\theta d\theta$. Now, we have $$\int \sec^2(2\theta) \tan(2\theta) d\theta=\frac{1}{2}\int udu=\frac{1}{4} u^2+c\\ =\frac{1}{4} (\tan 2\theta)^2+c.$$