Answer
$\dfrac{1}{2} \ln|\dfrac{2 \theta}{7} +\dfrac{\sqrt {4\theta^2-49}}{7}|+C$
Work Step by Step
Plug $\theta=\dfrac{7}{2} \sec \theta \implies d \theta=\dfrac{7}{2} \sec\theta \tan \theta d \theta $
Then,
$\int \dfrac{\dfrac{7}{2} \sec\theta \tan \theta d \theta}{\sqrt{4(\dfrac{49}{4} \sec^2 \theta-49)}} = \dfrac{1}{2} \int \sec \theta d \theta$
Substitute $\sec \theta =\dfrac{2 \theta}{7}$
$ \dfrac{1}{2} \ln |\sec\theta + \tan \theta|+C=\dfrac{1}{2} \ln|\dfrac{2 \theta}{7} +\dfrac{\sqrt {4\theta^2-49}}{7}|+C$
