If $I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+Ta{n}^2\theta +2Tan\theta )d\theta ,$

$I=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+Ta{n}^2\theta +2Tan\theta )d\theta$

$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log {(1+Tan\theta )}^{2}d\theta$

$=2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+Tan\theta )d\theta$

$=2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log \left(1+Tan\left(\frac{\pi }{4}-\theta \right)\right)d\theta$

$=2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log \left(\frac{1+\tan \theta +1-\tan \theta }{1+\tan \theta }\right)d\theta$

$=2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log 2d\theta -2\underset{0}{\overset{\frac{\pi }{4}}{\int }}\log (1+\tan \theta )d\theta$

$2I=2\left[\frac{{\displaystyle \pi }}{{\displaystyle 4}}\log 2\right]$

$I=\frac{\pi }{4}\log 2$