The integral $$\int_{7\pi/4}^{7\pi/3} \sqrt{\tan^2 x} \ dx$$ is equal to :

$$I = \int_{7\pi/4}^{7\pi/3} \sqrt{\tan^2 x} \, dx$$

$$\sqrt{\tan^2 x} = |\tan x|$$

$$I = \int_{7\pi/4}^{2\pi} (-\tan x) \, dx + \int_{2\pi}^{7\pi/3} \tan x \, dx \text{}$$

$$[-\log \sec x]_{7\pi/4}^{2\pi} = -(\log \sec(2\pi) - \log \sec(7\pi/4)) \text{}$$

= $$-(0 - \log \sqrt{2}) = \log \sqrt{2} \text{}$$

$$[\log \sec x]_{2\pi}^{7\pi/3} = \log \sec(7\pi/3) - \log \sec(2\pi) \text{}$$

= $$\log 2 - 0 = \log 2 \text{}$$

$$I = \log \sqrt{2} + \log 2 \text{}$$

$$I = \log (2\sqrt{2}) \text{}$$