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)#.
