STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

put $x=\tan \theta$

$\therefore \frac{d x}{d \theta}=\sec ^{2} \theta \Rightarrow d x=\sec ^{2} \theta d \theta$

$\therefore I = 8\int\limits_0^{\pi /4} {\frac{{\log (1 + \tan \theta )}}{{1 + {{\tan }^2}\theta }}.} {\sec ^2}\theta d\theta $

$I = 8\int\limits_0^{\pi /4} {\log (1 + \tan \theta )} d\theta .....\left( i \right)$

$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + \tan \left( {\frac{\pi }{4} - \theta } \right)} \right]} d\theta $

$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + \frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right]} d\theta $

$ = 8\int\limits_0^{\pi /4} {\log \left[ {\frac{2}{{1 + \tan \theta }}} \right]} d\theta $

$ = 8\int\limits_0^{\pi /4} {\left[ {\log 2 - \log \left( {1 + \tan \theta } \right)} \right]} d\theta $

$I=8 .(\log 2)[x]_{0}^{\pi / 4}-8$

$ = \int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta $

$I=8 . \frac{\pi}{4} \cdot \log 2-I[\text { From equation }(i)]$

$\Rightarrow 2 I=2 \pi \log 2$