Evaluate: $\int \frac {\sec (x)}{\sec (x)+\tan (x)} dx$ Evaluate: $\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx$
My Attempt:
$$=\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx$$
$$=\int \dfrac {\sec^2 (x)}{\sec^2 (x)+\sec (x).\tan (x)}$$
 A: $$\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx =\int \sec (x)(\sec x - \tan x) dx 
 = \int \sec^2 x -\tan x \sec x dx \\= \tan x - \sec x + C$$
A: Or...if you are not too keen on working with trig reciprocal functions: Multiply top and bottom by $\cos x$ to arrive at $\int{\frac{dx}{1+\sin x}}$.
Then multiply top and bottom by $1-\sin x$ to arrive at $\int{\frac{1-\sin x}{\cos^2 x}}dx$. Now split the fraction and it becomes standard work. It may not be quicker but at least you are working with the more familiar $\sin/\cos$ functions...
A: Another approach might be the following:
$$\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx =\int \dfrac {\sec (x)(\sec (x)+\tan (x))}{(\sec (x)+\tan (x))^2} dx \to\begin{bmatrix}u=\sec (x)+\tan (x)\\ du=\sec(x)(\sec (x)+\tan (x))\,dx\end{bmatrix}\to\int\frac1{u^2}\,du=\boxed{-\frac1{\sec(x)+\tan(x)}+C}$$
This method works for any integral of the form 
$$\int \dfrac {\sec (x)}{(\sec (x)+\tan (x))^n} dx $$
with $n\in \mathbb{Z}$. 
A: Another approach
$$
\begin{align}
I&=\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx\\
&=\int \dfrac {dx }{1+\sin (x)} \\
&=\int \dfrac {dx}{\sin(\pi/2)+\sin (x)} \\
&=\int \dfrac {dx}{2\cos(\pi/4 -x/2)\sin (x/2+\pi/4)} \\
&=\int \dfrac {dx}{2\sin^2 (x/2+\pi/4)} \\
I&=- \cot(x/2+\pi/4) +K
\end{align}
$$
A: Another approach: 
$$\begin{align}\int\ \dfrac{\sec x}{\sec x+\tan x}\,\mathrm dx
&=\int\ \dfrac{\sec^2 x-\sec x\cdot\tan x}{\sec^2 x-\tan^2 x}\,\mathrm dx
\\&=\int \sec^2 x\,\mathrm dx-\int\ \sec x\cdot \tan x\,\mathrm dx\\
&=\boxed{\tan x-\sec x+C}\end{align}$$
