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[tex] f(x) = \sqrt{4-x^2} [/tex]

Find the arc length of the given function from x=0 to x=2.

I know that I am supposed to use this formula to solve for arclength:

[tex] \int_{0}^{2} \sqrt{1 + (f\prime(x))^2} dx[/tex]

Ok so I take the derivative of

[tex] f(x) = \sqrt{4-x^2} [/tex]

and I get:

[tex] f\prime(x) =\frac{-x}{\sqrt{4-x^2}} [/tex]

I then plug this into the equation get:

[tex] \int_{0}^{2} \sqrt{1 + (\frac{-x}{\sqrt{4-x^2}}) ^2} [/tex]

[tex] \int_{0}^{2} \sqrt{1 + \frac{x^2}{4-x^2}} [/tex]

I get stuck here because I can't integrate this. I can't use any integral tables and I even try some algebraic manipulation inside of the square root and get:

[tex] \int_{0}^{2} \sqrt{\frac{4}{4-x^2}} [/tex]

Any help? Thank you.