Question

$\int \frac{\sec x}{\log (\sec x+\tan x)}dx$

• A. $\log \left[\log (\sec x-\tan x)\right].$
• B. $\log \left[\log (\sec x+\tan x)\right].$
• C. $\log \left[\log (\sin x+\cos x)\right].$
• D. $\log \left[\log (\sin x-\cos x)\right].$

Answer 1

The correct option is B $\log \left[\log (\sec x+\tan x)\right].$

Let $I=\int \frac{\sec x}{\log (\sec x+\tan x)}dx$
Put $\log (\sec x+\tan x)=t\Rightarrow \sec xdx=dt$
Therefore
$I=\int \frac{dt}{t}=\log t=\log \left[\log (\sec x+\tan x)\right]$
Hence, option 'B' is correct.