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

Question

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

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Answer 1 · 2018-06-13T02:31:27

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

#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))#