Math Is Fun Forum

I'm trying to calculate the curvlenght for y=ln(cosx) , 0<=x<=pi/6. But I can not figure out what I'm doing wrong.

Using sec(x) is not allowed (because we don't learn that here in Sweden).

Using formula ds = squareroot(1 + (dy/dx)^2) dx. dy/dx=-sinx/cosx

squareroot( 1 + (dy/dx)^2 ) dx = squareroot( 1 + (-sinx/cosx)^2 ) dx = squareroot( 1 + (tanx)^2 ) dx =

// 1+tan^2(x) = 1/cos^2(x) //

= squareroot( 1/cos^2(x) ) dx = ( squareroot(1) ) / ( squareroot(cos^2(x) ) dx = 1/cosx dx =

// 1/cosx = cosx/cos^2(x) //

= cosx/cos^2(x) dx =

//cos^2(x) = 1- sin^2(x) //

= cosx/( 1 - sin^2(x) ) dx = 1/( 1 - sin^2(x) * cosx dx

// Using u = sinx , du = cosx dx //

= 1/( 1 - u^2 ) du = 1/( (1 - u)(1+u) ) du =

// 1/( (1 - u)(1+u) ) = ( A/(1-u) ) + ( B/(1+u) ) => A+Au+B-Bu = 1 => A+B=1 and A-B=0 <=> A=1/2 , B=1/2 //

1/( 1 - u^2 ) du = (1/2) * ( (1/(1+u) + 1/(1-u) ) du = (1/2) * ( ln|1+u| + ln|1-u| ) =

// ln(A) + ln(B) = ln(A*B) , (1+u)(1-u)=1-u^2 //

= (1/2) * ln|1-u^2| =

// changing back u=sinx //

= (1/2) * ln| 1-sin^2(x) | = (1/2) * ln| cos^2(x) | = ln|cosx|

ds = squareroot(1 + (dy/dx)^2) dx , 0<=x<=pi/6 <=> ln| cosx | 0<=x<=pi/6

But when calculating ln| cos(pi/6) | - ln|1| I get -0,14... The correct answer should be (1/2)*ln(3) which is about 0,5