Integral substitution u = sin x

du/dx = cos x

dx = 1/[cos x] du

x = 0, u = 0
x = 1/2, u = sin (1/2)

Integral becomes

Integral between 0 and sin (1/2) of u^(1/2) cos x (1/cos x) du

Integral between 0 and sin (1/2) of u^(1/2) du

[(2/3)u^(3/2)] evaluated between 0 and sin (1/2)

Not sure where your - sin x came from?