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Evaluate : $ \int_0^{\large\frac{\pi}{4}} \log(1+\tan x)dx $

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  • $ \int \limits_a^b f(x)dx=F(b)-F(a)$
  • $\int \limits_0^a f(x)dx=\int \limits_0^a f(a-x)dx$
  • $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A \tan B}$
  • $\tan (A-B)=\large\frac{\tan A-\tan B}{1+\tan A \tan B}$
Step 1:
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 (\large\frac{\pi}{4}$$-x)]\;dx$
Using $\tan (A-B)=\large\frac{\tan A-\tan B}{1+\tan A \tan B}$
$I=\int \limits_0^{\frac{\pi}{4}} \log \bigg[1+\large\frac{\tan {\pi}{4}-\tan x}{1+\tan \frac{\pi}{4}.\tan x}\bigg]dx $
$I=\int \limits_0^{\frac{\pi}{4}} \log \bigg[1+\large\frac{1-\tan x}{1+\tan x}\bigg]dx $
$=\int \limits_0^{\frac{\pi}{4}} \log \large\frac{2}{(1+\tan x)}$$dx $
$\log(\frac{a}{b})=\log a-\log b$ similarly
$I=\int \limits_0^{\frac{\pi}{4}}\log 2dx-\int \limits_0^{\frac{\pi}{4}}\log(1+\tan x)dx$
$=\int \limits_0^{\frac{\pi}{4}} \log 2.dx-I$
Step 2:
Therefore $2I=\int \limits_0^{\frac{\pi}{4}} \log 2.dx$
On integrating we get,
$ 2I=[\log 2 (x)]_0^{\large\frac{\pi}{4}}$
Applying limits we get,
$2I=\log 2. (\pi/4)$
$=>2I=\large\frac{\pi}{4}$$\log 2$
Therefore $ I=\large\frac{\pi}{8}$$\log 2$
answered Nov 14, 2013 by sreemathi.v
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