$\int \frac{1}{\operatorname{cosec} x+\cos x} d x$
$\begin{aligned} & =\frac{1}{2 \sqrt{3}} \log \mid f(x) 1-\int \frac{\cos x-\sin x}{2+\sin 2 x} d x+C \\ & \Rightarrow \int \frac{d x}{\frac{1}{\sin x}+\cos x} \\ & =\int \frac{\sin x d x}{1+\sin x \cdot \cos x}=\int \frac{2 \sin x}{2+\sin 2 x} d x \\ & =\int \frac{\sin x+\cos x+\sin x-\cos x}{2+\sin 2 x} d x \\ & =\int \frac{\sin x+\cos x}{2+\sin 2 x}-\int \frac{\cos x-\sin x}{2+\sin 2 x} d x \\ & \therefore \quad \frac{1}{2 \sqrt{3}} \log |f(x)|=\int \frac{\sin x+\cos x}{2+\sin 2 x} d x \\ & I=\int \frac{\sin x+\cos x}{2+\sin 2 x} d x \\ & =\int \frac{\sin x+\cos x}{3-\sin ^2 x-\cos ^2 x+2 \sin x \cos x} d x\end{aligned}$
$=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^2} d x$
Let, $\sin x-\cos x=t,(\cos x+\sin x) d x=d t$
$\begin{aligned} & \int \frac{d t}{3-t^2}=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C \\ & =\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|+C \\ & \therefore \quad f(x)=\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x} \\ & \text { at } x=\frac{\pi}{3} \\ & \therefore \quad f(x)=\frac{\sqrt{3}+\frac{\sqrt{3}}{2}-\frac{1}{2}}{\sqrt{3}-\frac{\sqrt{3}}{2}+\frac{1}{2}}=\frac{3 \sqrt{3}-1}{\sqrt{3}+1} .\end{aligned}$