Question

The value of integral $\underset{0}{\overset{\theta }{\int }}\ln (1+\tan \theta \tan x)dx,\theta \in (0,\frac{\pi }{2})$ is equal to

• A. $2\theta \cdot \ln \sec \theta$
• B. $\theta \cdot \ln \sec \theta$
• C. $2\theta \cdot \ln \cos \theta$
• D. $\theta \cdot \ln \cos \theta$

Answer 1

The correct option is B $\theta \cdot \ln \sec \theta$

$I=\underset{0}{\overset{\theta }{\int }}\ln (1+\tan \theta \tan x)dx$
$=\underset{0}{\overset{\theta }{\int }}\ln (1+\tan \theta \tan (\theta -x\left)\right)dx=\underset{0}{\overset{\theta }{\int }}\ln (1+\tan \theta \left(\frac{\tan \theta -\tan x}{1+\tan \theta \tan x}\right))dx=\underset{0}{\overset{\theta }{\int }}\ln \left(\frac{1+{\tan }^2\theta }{1+\tan \theta \tan x}\right)dxI=\underset{0}{\overset{\theta }{\int }}\ln \left({\sec }^2\theta \right)dx-\underset{0}{\overset{\theta }{\int }}\ln (1+\tan \theta \tan x)dxI=2\theta \ln \sec \theta -I\Rightarrow I=\theta \cdot \ln \sec \theta$