Correct option: A.$\frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}\tan \frac{x}{2}} \right) + c$
a
(a)$I = \frac{{dx}}{{7 + 5\cos x}}$$ = \int {\frac{{dx}}{{7 + 5\,\left( {\frac{{1 - {{\tan }^2}(x/2)}}{{1 + {{\tan }^2}(x/2)}}} \right)}}} $
$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{7 + 7{{\tan }^2}(x/2) + 5 - 5{{\tan }^2}(x/2)}}} $
$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{12 + 2{{\tan }^2}(x/2)}}} $$ = \int {\frac{{\frac{1}{2}{{\sec }^2}(x/2)\,.dx}}{{6 + {{\tan }^2}(x/2)}}} $
$\tan \frac{x}{2} = t$ $⇒$ $\frac{1}{2}{\sec ^2}\frac{x}{2}dx = dt$
$I = \int {\frac{{dt}}{{{t^2} + ({{\sqrt {6)} }^2}}}} $ $ = \frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\frac{t}{{\sqrt 6 }} + c$
$ = \frac{1}{{\sqrt 6 }}{\tan ^{ - 1}}\left| {\frac{{\tan (x/2)}}{{\sqrt 6 }}} \right| + c$