IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

seraphic

$\int$ [1-x^1/2]^1/2/[1+x^1/2]^1/2 dx

= $\int$ [1- $\sqrt{x}$ ]/ $\sqrt{1-x}$

take $\sqrt{x}$ = t

= $\int$ (1-t)(2t)/ $\sqrt{1-}{t}^{2}$ dt= 2 $\int$ t/ $\sqrt{1-}{t}^{2}$ - t²/ $\sqrt{1-}{t}^{2}$ dt

=2/(-2) $\int$ -2t/ $\sqrt{1-}{t}^{2}$ dt + 2 $\int$ {(1-t²)/ $\sqrt{1-}{t}^{2}$ - 1/ $\sqrt{1-}{t}^{2}$ }dt

=-2 $\sqrt{1-}{t}^{2}$ + 2 $\int$ $\sqrt{1-}{t}^{2}$ - tan^-1(t)

= -2 $\sqrt{1-}{t}^{2}$ + 2[t/2 $\sqrt{1-}{t}^{2}$ + 1/2 sin^-1(t) - tan^-1(t)]

substitute t= $\sqrt{x}$