Question
$\int\limits_0^1 {\cfrac{{dx}}{{\sqrt {1 - {x^2}} }}}$
$\int\limits_0^1 {\cfrac{{dx}}{{\sqrt {1 - {x^2}} }}}$
: $\int\limits_0^1 {\cfrac{{dx}}{{\sqrt {1 - {x^2}} }}} = \left[ {{{\sin }^{ - 1}}x} \right]_0^1 = {\sin ^{ - 1}}\left( 1 \right) - {\sin ^{ - 1}}\left( 0 \right) = \cfrac{\pi }{2}$
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Integrals. Curated by Sachin Sharma. Free for all students.