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f[ (sinx)^2] = x/sinx
let
x= (siny)^2
dx = 2siny.cosy dy
∫√x/√(1-x) f(x) dx
=∫ [siny/cosy] f[(siny)^2] . [ 2siny.cosy dy ]
=2∫ (siny)^2. f[(siny)^2] dy
=2∫ (siny)^2 . [y/siny] dy
= 2∫ ysiny dy
= -2∫ ydcosy
= -2ycosy +2∫ cosy dy
= -2ycosy +2siny + C
= -2[arcsin(√x)] .√(1-x) +2√x + C
where
x= (siny)^2
y =arcsin(√x)
= arcos √(1-x)
cosy =√(1-x)
let
x= (siny)^2
dx = 2siny.cosy dy
∫√x/√(1-x) f(x) dx
=∫ [siny/cosy] f[(siny)^2] . [ 2siny.cosy dy ]
=2∫ (siny)^2. f[(siny)^2] dy
=2∫ (siny)^2 . [y/siny] dy
= 2∫ ysiny dy
= -2∫ ydcosy
= -2ycosy +2∫ cosy dy
= -2ycosy +2siny + C
= -2[arcsin(√x)] .√(1-x) +2√x + C
where
x= (siny)^2
y =arcsin(√x)
= arcos √(1-x)
cosy =√(1-x)
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