If \(\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x=\mathrm{A} \tan ^{-1}(\mathrm{~B} \cos x)+\mathrm{C}\), (where \(\mathrm{c}\) is a constant of integration), then the value of \(\mathrm{A}+\mathrm{B}\) is

Let \(\mathrm{I}=\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x\)
Put \(\cos x=\mathrm{t}\)
\(-\sin x \mathrm{~d} x=\mathrm{dt}\)
\(\therefore \quad \sin x \mathrm{~d} x=-\mathrm{dt}\)
\(\begin{aligned}
\therefore \quad \mathrm{I} & =\int \frac{-\mathrm{dt}}{3+4 \mathrm{t}^2} \\
& =-1 \int \frac{1}{4 \mathrm{t}^2+3}
\end{aligned}\)
\(=-1 \int \frac{1}{(2 t)^2+(\sqrt{3})^2}\)
\(=-1 \times \frac{1}{2 \times \sqrt{3}} \tan ^{-1}\left(\frac{2 \mathrm{t}}{\sqrt{3}}\right)+\mathrm{c}\)
\(\mathrm{I}=\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+\mathrm{c}\)
But \(\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x=\mathrm{A} \tan ^{-1}(\mathrm{~B} \cos x)+\mathrm{c}\)
Comparing above equations, we get
\(\begin{aligned}
\mathrm{A} & =\frac{-1}{2 \sqrt{3}}, \mathrm{~B}=\frac{2}{\sqrt{3}} \\
\therefore \quad \mathrm{A} & +\mathrm{B}=\frac{-1}{2 \sqrt{3}}+\frac{2}{\sqrt{3}}=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}
\end{aligned}\)