Q:

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...

