Question
$\int_0^a {\frac{{x\,dx}}{{\sqrt {{a^2} + {x^2}} }}} = $
$ \Rightarrow 2xdx = dt$
$\int_0^a {\frac{{xdx}}{{\sqrt {{a^2} + {x^2}} }} = \frac{1}{2}\int_{{a^2}}^{2{a^2}} {\frac{1}{{\sqrt t }}dt} } $
$ = [{(2{a^2})^{1/2}} - {a^{2/2}}] = a(\sqrt 2 - 1)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.