7.2 definite integral — ગણિત ધોરણ 12 સાયન્સ — Question

Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$

$\therefore $ $I = \int_0^{\pi /2} {\log {{(\sec \theta )}^2}d\theta = 2\int_0^{\pi /2} {\log \sec \theta \,\,d\theta } } $

$ = - 2\int_0^{\pi /2} {\log \cos \theta \,\,d\theta = - 2.\,\,\frac{\pi }{2}\log \frac{1}{2}} $

$ = - \pi \log \frac{1}{2} = \pi \log 2$.