Correct Answer - Option 4 : log [log(tan x)] + c
Concept:
sin 2x = 2sin x cos x
∫(1/x)dx = log x + c
∫tanx dx = sec2x + c
Calculation:
Let I = \(\smallint \frac{2}{{\sin 2{\rm{x}}.\log \left( {{\rm{tanx}}} \right)}}\) ....(1)
Take log (tan x) = t
\(\rm \frac1 {\tan x}(se{c^2}x)dx = dt\)
⇒ \( \frac{{{\rm{cosx}}}}{{{\rm{sinx}}.{\rm{\;co}}{{\rm{s}}^2}{\rm{x}}}}{\rm{\;dx}} = {\rm{dt}}\)
⇒ \( \frac{1}{{{\rm{sinx}}.{\rm{cosx}}}}{\rm{dx}} = {\rm{dt}}\)
⇒ dx = sin x.cos x dt
Putting the value of log (tan x) and dx in equation (i)
Now, I = \(\rm \smallint \frac{2}{{2sinx.cosx.t}}\;sinx.cosx\;dt\)
= ∫ \(\rm \frac 1 t\)dt
= log t + c
= log [log(tan x) ]+ c