How do you find the arc length of the curve #y=ln(cosx)# over the interval #[0,π/4]#?

1 Answer
Feb 25, 2015

The answer is: #ln(sqrt2+1)#.

The lenght of a function written in cartesian coordinates is:

#L=int_a^bsqrt(1+[f'(x)]^2)dx#.

So:

#y'=1/cosx*(-sinx)#

Than:

#L=int_0^(pi/4)sqrt(1+sin^2x/cos^2x)dx=int_0^(pi/4)sqrt((cos^2x+sin^2x)/cos^2x)dx=#

#=int_0^(pi/4)sqrt(1/cos^2x)dx=int_0^(pi/4)1/cosxdx=(1)#

Now we have to make a substituition (with tangent half-angle formulae):

#t=tan(x/2)rArrx/2=arctantrArrx=2arctanxrArr#

#dx=2/(1+t^2)dt#,

and

if #x=0# than #t=tan0=0#

if #x=pi/4# than, for the half angle-formula,
#t=tan(pi/8)=sin(pi/4)/(1+cos(pi/4))=(sqrt2/2)/(1+sqrt2/2)=#

#=(sqrt2/2)/((2+sqrt2)/2)=sqrt2/2*2/(2+sqrt2)=sqrt2/(2+sqrt2)=#

#sqrt2/(2+sqrt2)*(2-sqrt2)/(2-sqrt2)=(2sqrt2-2)/(4-2)=(2(sqrt2-1))/2=#

#=sqrt2-1#

So, remembering that #cosx=(1-t^2)/(1+t^2)#:

#(1)=int_0^(sqrt2-1)1/((1-t^2)/(1+t^2))*2/(1+t^2)dt=#

#=2int_0^(sqrt2-1)1/(1-t^2)dt=(2)#

#1/((1-t)(1+t))=A/(1-t)+B/(1+t)=(A(1+t)+B(1-t))/((1-t)(1+t))#

and two polynomials (#1# on the left and #[A(1+t)+B(1-t)]# on the right) are identical if they assume the same values at the same values of #t#:

If #t=1# then #1=2ArArrA=1/2#;

If #t=-1# then #1=2BrArrB=1/2#.

So:

#(2)=2int_0^(sqrt2-1)((1/2)/(1-t)+(1/2)/(1+t))dt=#

#=2*1/2int_0^(sqrt2-1)(1/(1-t)+1/(1+t))dt=#

#=int_0^(sqrt2-1)(-(-1)/(1-t)+1/(1+t))dt=[-ln|1-t|+ln|1+t|]_0^(sqrt2-1)=#

#=[-ln|1-(sqrt2-1)|+ln|1+(sqrt2-1)|-(ln|1-0|+ln|1+0|]=#

#=-ln(2-sqrt2)+ln(sqrt2)=ln(sqrt2/(2-sqrt2))=#

#=ln((sqrt2)/(2-sqrt2)*(2+sqrt2)/(2+sqrt2))=#

#=ln((2sqrt2+2)/(4-2))=ln((2(sqrt2+1))/2)=ln(sqrt2+1)#.