Make cosx = cos？ y
-sinx dx = -2siny comfortable dy
dx = 2 siny cosy dy/√( 1-cos^4(y))= 2 cosy dy/√( 1+cos？ y)
∫√( cosx)dx =∫cosy * 2 cosy/[√( 1+cos？ y)] dy
= 2∫ cos？ y/√( 1 + cos？ y) dy
= 2∫ √( 1 + cos？ y) dy - 2∫ dy/√( 1 + cos？ y)
= 2∫ √(2 - sin？ y) dy - 2∫ dy/√(2 - sin？ y)
= 2√2 ∫ √( 1 - 1/2 sin？ y)dy-2/√2∫dy/√( 1- 1/2 sin？ y)
= 2 √ 2e (1√ 2, k)-√ 2f (1√ 2, k), with a lower limit of 0 and an upper limit of k.
F(a, b) is the first kind of incomplete elliptic integral.
E(a, b) is the second kind of incomplete elliptic integral.
But the original functions of √tanx and √cotx are still elementary functions, which can be found.