Shibashis Mallik
Last Activity: 10 Years ago
∫(sinx)1/2dx = ∫(√2tanx/2)/(√1+tan2x/2) = √2 ∫ √tanx/2. secx/2 dx
(Let √tan x/2 = t => dx = (4dt√tanx/2)/sec2x/2 )
=4√2 ∫(√t secx/2 √t)/ sec2x/2 dt = 4√2 ∫tdt / √1+t2
(Let 1+t2=u => tdt=du/2)
= 4√2 * ½ ∫du/√u = 2√2 ∫ u-1/2 du = 2√2* 2√u = 4√2 * √1+t2 = 4√2*√1+tanx/2