7.2 definite integral — Maths STD 12 Science — Question

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

$ \Rightarrow I = \int_0^{\pi /2} {\,\,\,\,\,\log (\tan \theta + \cot \theta } )\frac{{{{\sec }^2}\theta }}{{{{\sec }^2}\theta }}\,d\theta $

==> $I = \int_0^{\pi /2} {\,\,\,\,\,\log (\tan \theta + \cot \theta } )d\theta $

$ \Rightarrow I = \int_0^{\pi /2} {\log \frac{{(1 + {{\tan }^2}\theta )}}{{\tan \theta }}\,d\theta } $

==> $I$ $ = 2\int_0^{\pi /2} {\log \sec \theta \,d\theta - \int_0^{\pi /2} {\log \tan \theta } } \,d\theta $

==> $I$ $ = 2\int_0^{\pi /2} {\log \sec \theta \,\,d\,\theta } $;

$\left\{ \,\because \int_{0}^{\pi /2}{\log \tan \theta =0} \right\}$

$ \Rightarrow \,I = - 2\int_0^{\pi /2} {\,\,\,\,\,\log \cos \theta \,d\theta } $

==>$I = - 2 \times \frac{{ - \pi }}{2}\log 2$,$\left\{ \because \int_{0}^{\pi /2}{\log \cos \theta =-\frac{\pi }{2}\log 2} \right\}$

==> $I = \pi \log 2$.