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Evaluate the integral by completing the square and applying appropriate formulas from geometry.
int_0^8 sqrt(8x-x^2)

1 Answer
Narad T.
Dec 19, 2017

The answer is #=8pi#

Explanation:

Let's start by completing the square

#8x-x^2=16-(16-8x+x^2)=16-(4-x)^2#

#=16-(x-4)^2#

Perform the substitution

#u=x-4#, #=>#, #du=dx#

Therefore,

#intsqrt(8-x^2)dx=intsqrt(16-u^2)du#

Let,

#u=4sintheta#, #=>#, #du=4costhetad theta#

#16-u^2=16-16sin^2theta=16cos^2theta#

#sqrt(16-u^2)=4costheta#

Thereforre,

#intsqrt(8-x^2)dx=int4costheta*4costhetad theta=16intcos^2theta d theta#

#cos2theta=2cos^2theta-1#

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

#intsqrt(8-x^2)dx=8int(1+cos2theta)d theta#

#=8theta+4(sin2theta)#

#=8arcsin(u/4)+4*2*u/4*sqrt(16-u^2)/4)#

#=8arcsin(u/4)+u/2sqrt(16-u^2))#

#=8arcsin((x-4)/4)+((x-4))/2sqrt(16-(x-4)^2)+C#

#=8arcsin((x-4)/4)+((x-4))/2sqrt(8x-x^2)+C#

So,

#int_0^8sqrt(8-x^2)dx=[8arcsin((x-4)/4)+((x-4))/2sqrt(8x-x^2)]_0^8#

#=(8arcsin((8-4)/4)+((8-4))/2sqrt(8*8-8^2))-(8arcsin((0-4)/4)+((0-4))/2sqrt(0))#

#=8arcsin(1)+0-8arcsin(-1)#

#=8*pi/2+8*pi/2#

#=8pi#

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