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).
