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Find the \(\smallint \frac{2}{{{\rm{sin}}2{\rm{x}}.{\rm{log}}\left( {{\rm{tanx}}} \right)}}\)
1. log (sin x) + c
2. log (cos x) + c
3.  log (tan x) + c
4.  log [log(tan x)] + c

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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

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