Verified answer

remember identities.

sin(u/2)= (1/√2)√(1-cosu)

So 1-sinx = 1-cos(π/2-x)

So √(1-sinx) = √[1-cos(π/2-x)] = √2*√[1-cos(π/2-x)]/√2

=√2*cos(π/4-x/2)

So we get 5√2*INTEGRAL [cos(π/4-x/2) dx]

5√2*(-2)sin(π/4-x/2)+C

= -10√2*sin(π/4-x/2)+C

Expand it if you want

√(1 - sin(x))√(1 + sin(x)) / √(1 + sin(x))

√((1 - sin²(x)) / √(1 + sin(x))

|cos(x)| / √(1 + sin(x))

Assuming an integral taken over x such that cos(x) ≥ 0,

cos(x) / √(1 + sin(x))

u = 1 + sin(x)

du = cos(x)