int_0^a frac{x , dx}{sqrt{a^2 + x^2}} = | vedclass

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(A) Let $I = \int_0^a \frac{x \, dx}{\sqrt{a^2 + x^2}}$.Substitute $t = a^2 + x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$.Wh

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$a(\sqrt{2} - 1)$

Vedclass · Answer

(A) Let $I = \int_0^a \frac{x \, dx}{\sqrt{a^2 + x^2}}$.Substitute $t = a^2 + x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{1}{2} \, dt$.When $x = 0$,$t = a^2 + 0^2 = a^2$.When $x = a$,$t = a^2 + a^2 = 2a^2$.Now,the integral becomes:$I = \int_{a^2}^{2a^2} \frac{1}{2\sqrt{t}} \, dt = \frac{1}{2} \int_{a^2}^{2a^2} t^{-1/2} \, dt$$I = \frac{1}{2} [2t^{1/2}]_{a^2}^{2a^2} = [\sqrt{t}]_{a^2}^{2a^2}$$I = \sqrt{2a^2} - \sqrt{a^2} = a\sqrt{2} - a = a(\sqrt{2} - 1)$.

$\int_0^a \frac{x \, dx}{\sqrt{a^2 + x^2}} = $

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