Question

What is the arc length of `f(x)=sqrt(x+2)` on `x in [0,2]`?

I know that the integral is complex but I need it to be solved using trigonometric substitution

Answer 1

`L=1/2(2sqrt17-3sqrt2)+1/4ln((4+sqrt17)/(3+2sqrt2))` units.

Explanation:

`f(x)=sqrt(x+2)`

`f'(x)=1/(2sqrt(x+2))`

Arc length is given by:

`L=int_0^2sqrt(1+1/(4x+8))dx`

Apply the substitution `4x+8=u^2`:

`L=1/2int_(2sqrt2)^4sqrt(u^2+1)du`

Apply the substitution `u=tantheta`:

`L=1/2intsec^3thetad theta`

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

`L=1/4[secthetatantheta+ln|sectheta+tantheta|]`

Reverse the substitution:

`L=1/4[usqrt(u^2+1)+ln|u+sqrt(u^2+1)|]_(2sqrt2)^4`

Hence

`L=1/2(2sqrt17-3sqrt2)+1/4ln((4+sqrt17)/(3+2sqrt2))`