$\int {\frac{{\sin x}}{{3 + 4{{\cos }^2}x}}} dx =$..........

Let $I = \int {\frac{{\sin x}}{{3 + 4{{\cos }^2}x}}} dx$

Let's put $\cos x = t \Rightarrow - \sin xdx = dt$

therefore,$I = - \int {\frac{{dt}}{{3 + 4{t^2}}}} = - \frac{1}{4}\int {\frac{{dt}}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2} + {t^2}}}}$

$= - \frac{1}{4} \cdot \frac{2}{{\sqrt 3 }}{\tan ^{ - 1}}\frac{{2t}}{{\sqrt 3 }} + C$

$= - \frac{1}{{2\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{2\cos x}}{{\sqrt 3 }}} \right) + C$