$\int \frac{\sec x ~cosec ~x}{\log \tan x} d x$
Answer & explanation
Correct answer: option 3
put log tan x = t
$\frac{1}{\tan x} \sec ^2 x d x=d t$
sec x cosec x dx = dt
$\int \frac{dt}{t}$ = log t + c = log (log tan x)
Hence (3) is the correct answer.