$\int \frac{1}{\sin x+\sec x}dx=$

$I=\int \frac{1}{\sin x+\sec x}dx=\int \frac{\cos x}{\sin x.\cos x+1}dx$

$=\frac{1}{2}\int \frac{(\cos x+\sin x)+(\cos x-\sin x)}{\sin x\cos x+1}dx$

$=\frac{1}{2}\{\int \frac{\sin x+\cos x}{\sin x\cos x+1}dx+\int \frac{\cos x-\sin x}{\sin x\cos x+1}dx\}$

$=\frac{1}{2}[\int \frac{2(\sin x+\cos x)dx}{3-{(\sin x-\cos x)}^2}+\int \frac{2(\cos x-\sin x)dx}{1+{(\sin x+\cos x)}^2}]$

Put $\sin x-\cos x=y$ and $\sin x+\cos x=z$

Then $(\cos x+\sin x)dx=dy$ and $(\cos x-\sin x)dx=dz$

$I=\int \frac{dy}{3-y^2}+\int \frac{dz}{1+z^2}=\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{3}+y}{\sqrt{3}-y}\right|+{\tan }^{-1}z+c$

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