Correct Answer - B
`int sin x log (tan x) dx`
`= - cos x log tan x - int (-cos x) (1)/(tan x). Sec^(2) x dx`
`= - cos x log tan x + int (1)/(sin x) dx`
`= - cos x log (tan x) + int (1 + "tan"^(2) (x)/(2))/(2 "tan "(x)/(2)) dx`
Let `t = "tan"(x)/(2)`
`rArr (dx)/(dt) = (2)/(1 + t^(2)) rArr dx = (2)/(1 + t^(2)).dt`
So, `- cos x. log (tan x) + int (1 + "tan"^(2) (x)/(2))/(2 "tan"(x)/(2)).dx`
`= - cos x. log (tan x) + int (1 + t^(2))/(2t).(2)/(1 + t^(2))dt`
`= - cos x log tan x + int (1)/(t) .dt`
`= - cos x log tan x + log (t) + c`
`= - cos x log tan x + log tan ((x)/(2)) + c`
