Come si integra #sqrt (1-x ^ 2) #?

lasciare #x=sintheta#, #=>#, #dx=costhetad theta#

#costheta=sqrt(1-x^2)#

#sin2theta=2sinthetacostheta=2xsqrt(1-x^2)#

Pertanto, l'integrale è

#I=intsqrt(1-x^2)dx=intcostheta*costheta d theta#

#=intcos^2thetad theta#

#cos2theta=2cos^2theta-1#

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

Perciò,

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

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

#=1/2arcsinx+1/2xsqrt(1-x^2)+C#