Explanation

(c)
Let $I = \int_{}^{}\frac{\sin{x\ }}{3 + 4\cos^{2}x}dx$
Put $\cos{x = t\ \Rightarrow \ - \sin{x\ dx = dt}}$
$\therefore I = \int_{}^{}{\frac{- dt}{3 + 4t^{2}} = \frac{- 1}{4}\int_{}^{}\frac{dt}{t^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2}}}$
$\Rightarrow I = - \frac{1}{4 \bullet \frac{\sqrt{3}}{2}} \bullet \tan^{- 1}{\frac{t}{\left( \frac{\sqrt{3}}{2} \right)} + c}$
$= - \frac{1}{2\sqrt{3}}\tan^{- 1}{\frac{2t}{\sqrt{3}} + c}$
$\Rightarrow I = \frac{1}{2\sqrt{3}}\tan^{- 1}{\left( \frac{2\cos x}{\sqrt{3}} \right) + c}$