Question

Evaluate
$\to \bf \displaystyle\int ^{ \frac{\pi}{4} } _{0} \bf{log(1 + tanx)dx}$
​

Answer 1

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log(1 + tanx)dx}$

Let above expression be equation (1).

We know that :-

$\boxed{\rm \bf \displaystyle\int ^{ a} _{0} \bf f(x) \: dx= \bf \displaystyle\int ^{ a} _{0} \bf f(a - x) \: dx}$

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log(1 + tan( \frac{\pi}{4} - x))dx}$

We know :-

$\boxed { \bf tan(a - b) = \frac{tan \: a - tan \: b}{1 + (tan \: a \times tan \: b)} }$

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log \bigg(1 + \frac{tan \frac{\pi}{4} - tan \: x}{1 + (tan\frac{\pi}{4} \times tan \: x)} \bigg)dx}$

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log \bigg(1 + \frac{1- tan \: x}{1 + tan \: x} \bigg)dx}$

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log \bigg( \frac{1 + tan \: x + 1- tan \: x}{1 + tan \: x} \bigg)dx}$

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm{log \bigg( \frac{2}{1 + tan \: x} \bigg)dx}$

We know :-

$\boxed {\rm log( \dfrac{a}{b} ) = log \: a - log \: b }$

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm \bigg( log( 2) \: - log(1 + tan \: x) \bigg) dx$

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm log( 2) \: dx \: - \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm log(1 + tan \: x) dx$

From equation (1) :-

$\rm\longrightarrow \rm I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm log( 2) dx\: - I$

$\rm\longrightarrow \rm 2I = \displaystyle\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm log( 2) \: dx$

$\rm\longrightarrow \rm 2I = \displaystyle \rm log( 2)\int ^{ \rm \: \frac{\pi}{4} } _{0} \rm1 \: dx$

$\rm\longrightarrow \rm 2I = \displaystyle \rm log( 2) \bigg [ \: x \: \bigg]^{ \rm \: \frac{\pi}{4} } _{0}$

$\rm\longrightarrow \rm 2I = \displaystyle \rm log( 2) \: ( \frac{\pi}{4} - 0)$

$\rm\longrightarrow \rm I = \displaystyle \rm \frac{\pi}{8} \: log( 2)$