$$
\begin{aligned}
& \int \frac{d x}{16-7 \sin ^2 x}=\int \frac{\operatorname{cosec}^2 x d x}{16 \operatorname{cosec}^2 x-7} \\
& =\int \frac{\operatorname{cosec}^2 x d x}{16+16 \cot ^2 x-7}=\int \frac{\operatorname{cosec}^2 x d x}{16 \cot ^2 x+9}
\end{aligned}
$$

Let $\cot x=t,-\operatorname{cosec}^2 x d x=d t$
$$
\begin{aligned}
& =-\int \frac{d t}{16 t^2+9}=-\frac{1}{16} \int \frac{d t}{t^2+\left(\frac{3}{4}\right)^2} \\
& =-\frac{1}{16} \times \frac{1}{\left(\frac{3}{4}\right)} \tan ^{-1}\left(\frac{4 t}{3}\right)+C \\
& =\frac{-1}{12} \tan ^{-1}\left(\frac{4 \cot x}{3}\right)+C=\frac{1}{12} \tan ^{-1}\left(\frac{3 \tan x}{4}\right)+C .
\end{aligned}
$$