How do you find the integral of # sqrt (2x - x^2) dx#?

#2x-x^2=1-(x-1)^2# by completing the square.

Therefore, the integral is

#I=intsqrt(2x-x^2)dx=intsqrt(1-(x-1)^2)dx#

Let #u=x-1#, #=>#, #du=dx#

#I=intsqrt(1-u^2)du#

Let #u=sintheta#, #=>#, #du=costhetad theta#

#sqrt(1-u^2)=sqrt(1-sin^2theta)=costheta#

#I=intcostheta*costhetad theta=intcos^2theta d theta#

#cos2theta=2cos^2theta-1#

#=>#, #cos^2theta=(1+cos2theta)/2#

Therefore,

#I=1/2int(1+cos2theta)d theta#

#=1/2(theta+1/2sin2theta)#

#=1/2theta+1/4sin2theta#

#=1/2theta+1/2sinthetacostheta#

#=1/2arcsin(u)+1/2usqrt(1-u^2)#

#=1/2arcsin(x-1)+1/2(x-1)sqrt(1-(x-1)^2)+C#

#=1/2arcsin(x-1)+1/2(x-1)sqrt(2x-x^2)+C#

Here is Second Method :

Let, #x=2sin^2u. :. dx=2*2sinu*cosudu#.

#:. intsqrt(2x-x^2)dx#,

#=intsqrt(4sin^2u-4sin^4u)(4sinucosu)du#,

#=4intsqrt{4sin^2u(1-sin^2u)}sinucosudu#,

#=4int(2sinucosu)sinucosudu#,

#=2int(4sin^2ucos^2u)du#,

#=2intsin^2 2udu#,

#=2int(1-cos4u)/2du#,

#=u-1/4sin4u#,

#=u-1/4*2sin2ucos2u#,

#=u-1/2(2sinucosu)(1-2sin^2u)#,

#=u-sqrt{sin^2ucos^2u}(1-2sin^2u)#,

#=u-sqrt{sin^2u(1-sin^2u)}(1-2sin^2u)#.

Since, #x=2sin^2u, sinu=sqrt(x/2). :. u=arcsinsqrt(x/2)#.

# rArr intsqrt(2x-x^2)dx#,

#=arcsinsqrt(x/2)-sqrt{x/2(1-x/2)}(1-x)#,

#=arcsinsqrt(x/2)-1/2(1-x)sqrt(2x-x^2)+C#.

Enjoy Maths.!