Integrate sqrt(sinx) with limit 0 to pi/2.

∫√sin(x)dx

Rewrite/simplify using trigonometric/hyperbolic identities:

=∫√2cos^2(2x−π) / [4])−1dx

=∫√1−2sin^2(2x−π) / [4])dx

Substitute u=(2x−π) / [4])⟶ du/dx=1/ 2 ⟶ dx=2du:

=2∫√1−2sin^2(u)du

Now solving:

∫√1−2sin2(u)du

This is a special integral (incomplete elliptic integral of the second kind):

=E(u|2)

Plug in solved integrals:

2∫√1−2sin^2(u)du

=2E(u|2)

Undo substitution u=(2x−π) / [4]):

=2E(2x−π) / [4])∣2)

∫√sin(x)dx

=2E((2x−π) / [4])|2)+C

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