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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int \limits_0^{\large\frac{\pi}{4}} $$\log |1+ \tan x| dx$

\[\begin {array} {1 1} (a)\;\frac{\pi}{4} \log 2 \\ (b)\;\frac{\pi}{\theta} \log 6 \\ (c)\;\frac{\pi}{8} \log 2 \\ (d)\;none \end {array}\]

1 Answer

$f(x)= f(\large\frac{\pi}{4}$$-x)$
By using property
$=> f( \large\frac{\pi}{4} -x)= \int\limits_0^{\frac{\pi}{4}}$$ \log |1+ tan (z/4-x)|dx$
$=> f( \large\frac{\pi}{4} -x)= \int\limits_0^{\frac{\pi}{4}}$$ \log \large\frac{\{1+\tan x +1- \tan x\}}{1+\tan x}$$dx$
$ f( \large\frac{\pi}{4} -x)= \int\limits_0^{\frac{\pi}{4}}$$ \log |2| -\int \limits_0^{\large\frac{\pi}{4}} \log 2 dx$
$2f(x) =\log 2 [\large\frac{\pi}{4}$$-0]$
$f(x) =\large\frac{\pi}{8}$$ \log 2$
Hence c is the correct answer.
answered Dec 13, 2013 by meena.p
 
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