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By using the properties of definite integrals,evaluate the integral\[\int\limits_0^\frac{\pi}{4} log (1+\tan x)\;dx\]

$\begin{array}{1 1} \frac{\pi}{8} \log 2 \\ \frac{\pi}{4} \log 16 \\ \frac{\pi}{8} \log 8 \\ \frac{\pi}{4} \log 8 \end{array} $

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  • (i)$ \int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
  • (iii)$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$
  • (iv)$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$
Given I=$\int\limits_0^\frac{\pi}{4} log (1+\tan x)\;dx-----(1)$
By using the property $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
I=$\int\limits_0^\frac{\pi}{4} log [(1+\tan (\frac{\pi}{4}-x)]\;dx$
Using $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$
$I=\int \limits_0^{\frac{\pi}{4}} log \bigg[1+\frac{\tan {\pi}{4}-\tan x}{1+\tan \frac{\pi}{4}.\tan x}\bigg]dx $
$I=\int \limits_0^{\frac{\pi}{4}} log \bigg[1+\frac{1-\tan x}{1+\tan x}\bigg]dx $
$=\int \limits_0^{\frac{\pi}{4}} log \frac{2}{(1+\tan x)}dx $
$log(\frac{a}{b})=log a-log b$ similarly
$I=\int \limits_0^{\frac{\pi}{4}}log 2dx-\int \limits_0^{\frac{\pi}{4}}log(1+\tan x)dx$
$=\int \limits_0^{\frac{\pi}{4}} log 2.dx-I$
Therefore $2I=\int \limits_0^{\frac{\pi}{4}} log 2.dx$
On integrating we get,
$ 2I=[log 2 (x)]_0^{\frac{\pi}{4}}$
Applying limits we get,
$2I=\log 2. (\pi/4)$
$=>2I=\frac{\pi}{4}\log 2$
Therefore $ I=\frac{\pi}{8} log 2$



answered Feb 13, 2013 by meena.p

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