STD 12 - 7.2 definite integral — JEE कक्षा 11 साइन्स — Question

$\therefore $ $dx = {\sec ^2}\theta \,d\theta $

$x = 1 \Rightarrow \theta = \frac{\pi }{4}$ व $x = 0 \Rightarrow \theta = 0$, तो

$I = 2\int_0^{\pi /4} {\theta {{\sec }^2}\theta \,d\theta = 2[\theta \tan \theta ]_0^{\pi /4} - 2\int_0^{\pi /4} {\tan \theta \,d\theta } } $

$= \frac{\pi }{2} + 2\,[\log \cos x]_0^{\pi /4} = \frac{\pi }{2} - 2\log \sqrt 2 $.