How do you find the arc length of #y=ln(cos(x))# on the interval #[pi/6,pi/4]#?

1 Answer
Aug 21, 2014

You can find the Arc Length of a function by first finding its derivative and plugging into the known formula:

#L = int_a^bsqrt(1 + (dy/dx)^2)dx#

Process:

With our function of #ln(cos(x))#, we must first find its derivative. The shortcut for #ln(u)# -type functions is to just take the derivative of the inside and place it over the original of what's inside. Since the derivative of #cos(x)# is (#-sinx#), we end up with:

#dy/dx = -sinx/cosx#,

which is equal to:

#-tanx#.

Plugging into our Arc Length formula, we have:

#L = int_a^b sqrt(1 + (-tanx)^2)dx#.

If we square the #-tanx# term, we get:

#L = int_a^b sqrt(1 + tan^2(x))dx#

Since #1+tan^2(x) = sec^2(x)# is one of our known trig identities, we can change our equation into:

#L = int_a^b sqrt(sec^2(x))dx#, which simplifies to #L = int_a^b secx dx#

Now we must remember that #int secx# = #ln(secx + tanx) + C#, so for our equation we must solve:

#ln(secx + tanx)# from #pi/6# to #pi/4#, giving us:

#L = ln(2/sqrt2 + 1) - ln(2/sqrt3 + 1/sqrt3)#

#2/sqrt2# can be simplified to #sqrt2#, and the right term has a common denominator, which lets you add them together to become #3/sqrt3#, which is simplified#sqrt3#, giving us:

#L = ln(sqrt2 + 1) - ln(sqrt3)#

If you remember that #ln(a) - ln(b) = ln(a/b)#, we can simplify our answer to get:

#L = ln((sqrt2 + 1)/sqrt3)#

We can evaluate this for a decimal answer:

#L ~~ 0.332067...#