How to find $\int \frac {\sec x}{1+ \csc x} dx $? 
How to find $\int \frac {\sec x}{1+ \csc x} dx  $ ?

Well it reduces to $ \int \frac {\sin x}{\cos x (1+\sin x)} dx $ .Any hints next ?
I'm looking for a short and simple method without partial fractions if possibe.
 A: Hint:
$$ \int \frac {\sin x}{\cos x (1+\sin x)} dx =\int\frac{\sin x\cos x}{\cos^2x(1+\sin x)}dx=\int\frac{\sin x\cos x}{(1-\sin^2 x)(1+\sin x)}dx\int\frac{\sin x\cos x}{(1+\sin x)^2(1-\sin x)}dx$$
Take $u=\sin x$ and use partial fractions.
A: First, we write
$$
\frac{\sec x}{1+\csc x}=\frac{\tan x}{1+\sin x}=\frac{\tan x(1-\sin x)}{1-\sin^2x}=\frac{1}{\cos^2x}\tan x-\frac{1}{\cos x}\tan^2 x.
$$
Now
$$
\int\frac{1}{\cos^2x}\tan x\,dx=\frac{1}{2}\tan^2x.
$$
For the other part we integrate by parts,
$$
\begin{aligned}
\int\frac{1}{\cos x}\tan^2x\,dx&=\int\frac{\sin x}{\cos^2 x}\tan x\,dx\\
&=\frac{1}{\cos x}\tan x-\int\frac{1}{\cos x}(1+\tan^2x)\,dx\\
&=\frac{1}{\cos x}\tan x-\int\frac{\cos x}{1-\sin^2x}\,dx-\int\frac{1}{\cos x}\tan^2x\,dx\\
&=\frac{1}{\cos x}\tan x-\text{artanh}\,\sin x-\int\frac{1}{\cos x}\tan^2x\,dx.
\end{aligned}
$$
Now it is a matter of rearranging. I get the final result to be something like

$$ \int\frac{\sec x}{1+\csc x}\,dx=\frac{1}{2}\text{artanh}\,(\sin x)-\frac{1}{2}\frac{1}{\cos x}\tan x+\frac{1}{2}\tan^2x+C.$$

