Find the arc length of the curve

y=ln(cosx) from
x=0 to
x=π/6.
Length:

Well, that's a bit sneaky! You are asking me to find the "length" of a curve that is not exactly a straight line. Are you trying to plot a prank on me? Haha, good one!

But don't worry, I'm here to help. To find the arc length of the curve, we'll use a formula that involves integration. First, let's find the derivative of ln(cos(x)). The derivative is -tan(x).

Now, the formula for arc length is given by:

L = ∫(a to b) sqrt(1 + (dy/dx)^2) dx

Substituting the derivative we found earlier,

L = ∫(0 to π/6) sqrt(1 + tan^2(x)) dx

Now, instead of trying to directly calculate this integral, let's simplify it using a trigonometric identity. Recall the identity: sec^2(x) = 1 + tan^2(x).

The integrand becomes:

L = ∫(0 to π/6) sqrt(sec^2(x)) dx

Now, taking the square root of sec^2(x), we get:

L = ∫(0 to π/6) sec(x) dx

Integrating sec(x) gives us ln(sec(x) + tan(x)).

Finally, evaluating the integral from x = 0 to π/6:

L = ln(sec(π/6) + tan(π/6)) - ln(sec(0) + tan(0))

Now you can plug in the values and calculate the result. Remember, math is fun, even when it's a bit twisted!

To find the arc length of the curve y = ln(cos(x)) from x=0 to x=π/6, we need to use the arc length formula for a curve given by y = f(x):

L = ∫√(1 + (f'(x))^2) dx

First, let's find the derivative of y = ln(cos(x)):

y' = [(cos(x))^-1] * (-sin(x)) = - sin(x) / cos(x) = - tan(x)

Now we can substitute this derivative back into the arc length formula:

L = ∫√(1 + (-tan(x))^2) dx
L = ∫√(1 + tan^2(x)) dx
L = ∫√(sec^2(x)) dx
L = ∫sec(x) dx

To integrate sec(x), we can use the substitution method. Let u = sec(x), then du = sec(x)tan(x) dx:

L = ∫du
L = u + C

Now we need to find the antiderivative of sec(x):

∫sec(x) dx = ln|sec(x) + tan(x)| + C

Substituting back in:

L = u + C
L = ln|sec(x) + tan(x)| + C

Now we can evaluate this expression for x=0 and x=π/6:

L = ln|sec(π/6) + tan(π/6)| - ln|sec(0) + tan(0)|

Using the values of sec(π/6) = sqrt(3)/2, tan(π/6) = 1/√3, sec(0) = 1, and tan(0) = 0, we can simplify further:

L = ln|√3/2 + 1/√3| - ln|1 + 0|
L = ln|(3 + 2√3) / (2√3)| - ln(1)
L = ln|(3 + 2√3) / (2√3)|

Therefore, the arc length of the curve y = ln(cos(x)) from x=0 to x=π/6 is ln|(3 + 2√3) / (2√3)|.

To find the arc length of a curve, we need to use the formula for arc length:

L = ∫√(1 + (dy/dx)²) dx

First, let's find dy/dx by taking the derivative of y with respect to x:

Given y = ln(cosx), we can use the chain rule for differentiation. The derivative of ln(u) with respect to x is (1/u) * du/dx.

So, dy/dx = (1/cosx) * (-sinx) = -sinx/cosx = -tanx.

Now, we can substitute dy/dx into the arc length formula:

L = ∫√(1 + (-tanx)²) dx

To integrate this expression, we need to simplify the square root term:

1 + (-tanx)² = 1 + tan²x = sec²x

So, L = ∫√(sec²x) dx = ∫secx dx

To integrate secx, we can use the identity:

∫secx dx = ln|secx + tanx| + C

Now, we can evaluate the arc length from x = 0 to x = π/6:

L = ∫[0,π/6] secx dx = ln|sec(π/6) + tan(π/6)| - ln|sec(0) + tan(0)|

sec(π/6) = 2/√3, tan(π/6) = 1/√3
sec(0) = 1, tan(0) = 0

L = ln|(2/√3) + (1/√3)| - ln|1 + 0|
L = ln|(3/√3)| = ln(√3) = ln(√3)

Therefore, the length of the curve y = ln(cosx) from x = 0 to x = π/6 is ln(√3).

in general , arc length = ∫ √(1 + (dy/dx)^2) dx from x=a to x=b

for yours:
y = ln(cosx)
dy/dx = -sinx/cosx = -tanx

arc length = ∫ (1 + tan^2 x)^(1/2) dx from 0 to π/6
= ∫ (sec^2 x)^(1/2) dx from 0 to π/6
= ∫ secx dx from 0 to π/6
= | [ ln( (tanx) + secx) ] | from 0 to π/6

= | ln(tan(π/6) + sec π/6) - (ln((tan0) + sec 0) ) |
= | ln (1/√3 + 2/√3) - ln( 0 + 1) |
= | ln(3/√3) - 0 |
= l ln(3) - ln(√3) |
= ln 3 -(1/2)ln 3
= (1/2) ln 3

the integral of secx should be part of your repertoire of basic integrals, if not,
here is the derivation of ∫secx dx = ln(tanx) + secx

https://www.youtube.com/watch?v=d7bAxU5CDtk