Explanation

(a)
Let $I = \int_{}^{}\frac{dx}{7 + 5\cos x} = \int_{}^{}\frac{dx}{7 + 5\frac{1 - \tan^{2}\left( \frac{x}{2} \right)}{1 + \tan^{2}\left( \frac{x}{2} \right)}}$
$= \int_{}^{}\frac{\sec^{2}{\left( \frac{x}{2} \right)dx}}{7 + 7\tan^{2}{\left( \frac{x}{2} \right) + 5 - 5\tan^{2}\left( \frac{x}{2} \right)}}$
$= \int_{}^{}{\frac{\sec^{2}{\left( \frac{x}{2} \right)dx}}{12 + 2\tan^{2}\left( \frac{x}{2} \right)} = \int_{}^{}\frac{\frac{1}{2}\sec^{2}{\left( \frac{x}{2} \right)dx}}{6 + \tan^{2}\left( \frac{x}{2} \right)}}$
Put $\tan{\frac{x}{2} = t \Rightarrow \frac{1}{2}\sec^{2}{\frac{x}{2}dx = dt}}$
$\therefore I = \int_{}^{}{\frac{dt}{t^{2} + \left( \sqrt{6} \right)^{2}} = \frac{1}{\sqrt{6}}\tan^{- 1}{\left( \ \frac{t}{\sqrt{6}}\ \right) + c}}$
$= \frac{1}{\sqrt{6}}\tan^{- 1}{\left| \frac{\tan\left( \frac{x}{2} \right)}{\sqrt{6}} \right| + c}$