Question

Evaluate : $\int \frac{\sec xdx}{\log (\sec x+\tan x)}$

Answer 1

Let $I=\int \frac{\sec xdx}{\log (\sec x+\tan x)}$

Let, $\log (\sec x+\tan x)=z$

$⟹\sec xdx=dz$

So, $I=\int \frac{dz}{z}=\log z+c$

Hence $I=\log \left(\log (\sec x+\tan x)\right)+c$