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

Question

William Abdalla · Accepted Answer

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

#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#

Ezra Adams · Answer

# arcsinsqrt(x/2)-1/2(1-x)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.!

Axel Alvarez · Answer

undefined To find the integral of sqrt(2x - x^2) dx, you can use trigonometric substitution. Let x = (1 - cos(t)), then dx = sin(t) dt. Substitute these into the integral and simplify. You'll end up with an integral in terms of t. After integrating, convert back to x using the original substitution.