7.2 definite integral — Maths STD 12 Science — Question

Put $x = {\tan ^2}\theta \Rightarrow \theta = {\tan ^{ - 1}}\sqrt x $

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

$\therefore $$I = \int_{{{\tan }^{ - 1}}\sqrt 8 }^{{{\tan }^{ - 1}}\sqrt {15} } {\frac{{2\tan \theta {{\sec }^2}\theta \,}}{{({{\tan }^2}\theta - 3)\sqrt {{{\tan }^2}\theta + 1} }}d\theta } $

$ = \int_{{{\tan }^{ - 1}}\sqrt 8 }^{{{\tan }^{ - 1}}\sqrt {15} } {\frac{{2\tan \theta {{\sec }^2}\theta \,}}{{({{\sec }^2}\theta - 4)\sec \theta }}d\theta } $

$ = \int_{{{\tan }^{ - 1}}\sqrt 8 }^{{{\tan }^{ - 1}}\sqrt {15} } {\frac{{2\tan \theta \sec \theta }}{{({{\sec }^2}\theta - 4)}}\,d\theta } $

$ = \int_{{{\tan }^{ - 1}}\sqrt 8 }^{{{\tan }^{ - 1}}\sqrt {15} } {\frac{{2\tan \theta \sec \theta }}{{(\sec \theta - 2)(\sec \theta + 2)}}\,d\theta } $

$ = \left[ {\frac{1}{2}\log \frac{{(\sec \theta - 2)}}{{(\sec \theta + 2)}}} \right]_{{{\tan }^{ - 1}}\sqrt 8 }^{{{\tan }^{ - 1}}\sqrt {15} }$

$ = \frac{1}{2}\left[ {\log \frac{2}{6} - \log \frac{1}{5}} \right] $

$= \frac{1}{2}\log \frac{5}{3}$.