Nice and wonderful exercise shared by Dan Sitrau in RMM facebook group

Exercise:

Show that, $\int_{0}^{\gamma }{\ln \left( 1+\tan \gamma \tan x \right)dx}=\gamma \ln \sec \gamma $ with $-\frac{\pi }{2}\le \gamma \le \frac{\pi }{2}$

Solution: Let $I=\int_{0}^{\gamma }{\ln \left( 1+\tan \gamma \tan x \right)dx}=\int_{0}^{\gamma }{\ln \left( 1+\tan \gamma \tan \left( \gamma -x \right) \right)dx}$

(          using $\int_{a}^{b}{f\left( x \right)dx}=\int_{a}^{b}{f\left( a+b-x \right)dx}$ ……….(*)           )

But $\tan \left( \gamma -x \right)=\frac{\tan \gamma -\tan x}{1+\tan x\tan \gamma }\Leftrightarrow \tan \gamma \tan \left( \gamma -x \right)=\frac{-\tan x\tan \gamma +{{\tan }^{2}}\gamma }{1+\tan x\tan \gamma }$

Hence, $I=\int_{0}^{\gamma }{\ln \left( 1+\frac{-\tan x\tan \gamma +{{\tan }^{2}}\gamma }{1+\tan x\tan \gamma } \right)dx}=\int_{0}^{\gamma }{\ln \left( \frac{1+\tan x\tan \gamma -\tan x\tan \gamma +{{\tan }^{2}}\gamma }{1+\tan x\tan \gamma } \right)dx}$

$\Rightarrow I=\int_{0}^{\gamma }{\ln \left( 1+{{\tan }^{2}}\gamma  \right)dx-}\int_{0}^{\gamma }{\ln \left( 1+\tan x\tan \gamma  \right)dx}$

$\Rightarrow 2I=\int_{0}^{\gamma }{\ln \left( 1+{{\tan }^{2}}\gamma  \right)dx}=\int_{0}^{\gamma }{\ln \left( {{\sec }^{2}}\gamma  \right)dx}=2\gamma \ln \sec \gamma $

Therefore, $I=\int_{0}^{\gamma }{\ln \left( 1+\tan \gamma \tan x \right)dx}=\gamma \ln \sec \gamma $                               Q.E.D



(*) The idea of solution credit to the professor Ravi Prakash