How to solve: ∫(√(x))dx / √(1-x)

Let √(x) = sin(u)

Differentiate both sides

= 1/(2*√(x))dx = cos(u)du

So:

dx = cos(u)du  / 1/(2*√(x))

= cos(u)du / 1/[2*sin(u)]

Invert and multiply

 = cos(u)du*2*sin(u)

So:

∫√(x)dx / √(1-x)

Substitute

=  ∫sin(u ) * cos(u) * 2*sin(u)  / √(1-sin(u)^2) du

=  ∫ 2*sin(u)^2 * cos(u)  / √(1-sin(u)^2) du

Identity

=  ∫ 2*sin(u)^2 * cos(u)  / √(cos(u)^2) du

=  ∫2*sin(u)^2 * cos(u) / cos(u)du

cos(u) cancels

=  ∫2 * sin(u)^2 du

Identity

=  ∫2*[1 – cos(2u)]/2du

=  ∫ [1 – cos(2u)]du

=  ∫(1)du – ∫cos(2u)du

=  u – (1/2) * sin(2u) + C

Identity

u – (1/2)(2) * sin(u) * cos(u) + C

u – sin(u) * cos(u) + C

Back substitute

Answer:

 = sin-¹(√(x)) – √(x) *  √(1-x)”