Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 9

$\frac{1}{2}sec$ 2t+C

Putting u= 2t, we get $\frac{du}{dx}= 2 $ or dx=$ \frac{du}{2}$ Now, $\int$sec 2t tan 2t dt =$\int$ sec u tan u $ \frac{du}{2}$ =$\frac{1}{2}\int$sec u tan u du = $\frac{1}{2}\times sec $ u +C= $\frac{1}{2}sec$ 2t+C