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Find the exact length of the curve y = ln(sec(x)), 0 ≤ x ≤ 3?

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Final answer:

To find the exact length of the curve y = ln(sec(x)), we use the formula for arc length. We integrate the square root of 1 plus the square of the derivative of y with respect to x over the given interval. Using trigonometric identities and a substitution, we simplify the integral and determine the length.

Step-by-step explanation:

To find the exact length of the curve y = ln(sec(x)), 0 ≤ x ≤ 3, we will use the formula for arc length:

L = ∫[a, b] √(1 + (dy/dx)²) dx

In this case, dy/dx = d/dx(ln(sec(x))) = 1/(cos(x)sec(x)) = cos(x)

So the length of the curve is:

L = ∫[0, 3] √(1 + cos²(x)) dx

Using the trigonometric identity sin²(x) = 1 - cos²(x), we can simplify the integral:

L = ∫[0, 3] √(2 - sin²(x)) dx

Now we can use a trigonometric substitution u = sin(x) to further simplify the integral. We have:

du = cos(x) dx, so dx = du/cos(x)

L = ∫[0, 3] √(2 - u²) du/cos(x)

Since x ranges from 0 to 3, we need to express the integral in terms of u. The limits become 0 and sin(3). The integral becomes:

L = ∫[0, sin(3)] √(2 - u²) du/cos(x)

You can evaluate the integral and find the exact length of the curve.

User Flauschtrud
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