The answer is: #ln(sqrt2+1)#
To find the lenght of a curve #L#, written in cartesian coordinates, it is necessary to use this formula:
#L=int_a^bsqrt((1+[f'(x)]^2))dx#.
Since #f'(x)=1/cosx*(-sinx)#, then:
#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#.
This integral has to be done using this substitution (parametric formulae):
#t=tan(x/2)rArrx/2=arctantrArrx=2arctanxrArrdx=2/(1+t^2)dt#,
and it's known that: #cosx=(1-t^2)/(1+t^2)#,
if #x=0# then #t=0#
if #x=pi/4# then #t=tan(pi/8)=sqrt2-1#
So:
#int_0^(sqrt2-1)1/cosxdx=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#,
#1/(1-t^2)=1/((1+t)(1-t))=A/(1+t)+B/(1-t)=(A(1-t)+B(1+t))/((1+t)(1-t))#,
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=A*2rArrA=1/2#;
If #t=1# then #1=B*2rArrB=1/2#.
The integral becomes:
#2int_0^(sqrt2-1)[(1/2)/(1+t)+(1/2)/(1-t)]dt=2(1/2)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|=lnsqrt2-ln(2-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)#.
