High school mathematics: find definite integral, the interval is 0~ 1. Under the radical sign of the integrand function (2x-x 2). Well, I can't even type, and the root symbol. How to find the original function? Find the definite integral (0, 1) ∫ [√ (2x-x? )]dx

Solution: Original formula =. (0， 1) ∫ [√ (2x-x？ )]dx=(0， 1)√√[ 1-(x- 1)？ ]d(x- 1)

Let x- 1=sinu, then d(x- 1)=cosudu, when 0≤x≤ 1-1≤x- 1≤0, -π/2≤u≤0.

So the original formula = (-π/2,0) ∫ [√ (1-sin? u)]cosudu=(-π/2，0)∫cos？ udu

=(-π/2，0)∫[( 1+cos2u)/2]du=(-π/2，0)( 1/2)[∫du+( 1/2)∫cos2ud(2u)]

=(-π/2，0)( 1/2)[u+( 1/2)sin2u]=(u/2+( 1/4)sin2u)│(-π/2，0)=π/4