In Exercises 139–142, find the length of each curve.
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Question

In Exercises 139–142, find the length of each curve.

141. y = ln(cos(x)) from x = 0 to x = π/4.

Accepted Answer

Welcome back everyone. Find the length of the curve y equals g ln of cosine x from x equals 0 to x equals 5 divided by 6. A elon of 2 units, b 1/2 ln of 2 units, c 1/2 lon of 3 units, and d ln of 3 units. For this problem, let's remember that we find arc length L by using the integral from A to B of. The square root of 1 plus the derivative of y2 d x. So, let's begin by finding the derivative of why. The derivative of Y is going to be found by taking the derivative of negative ln of cosine x. We can factor out the negative sign, right, so that would be negative, and let's begin integrating ln of cosine x, understanding that it is a composite function. So we want to apply the chain rule, differentiating ln of cosine x with respect to cosine x. We get 1 divided by cosine x, and we want to multiply by the derivative of cosine x using the chain rule. So we get negative. The derivative of cosine x is negative sin x. And we're dividing by cosine x. So negative of negative gives us positive, and we get sin x divided by cosine x, which is basically tangent x. Now 1 plus the derivative of y. Squad would be 1 + tangent square x and using trigonometric identities that's sequin square x. If we take the square root of both sides. That be the square root of sequin square x, which is equal to the absolute value of sequent x. And specifically between 0 and 5 divided by 62 X is positive, so we're going to take the positive solution, right? Meaning arc length l is equal to the integral from 0 to pi divided by 6. These are the limits of integration based on the problem. And we're integrating sequent X D X. Let's remember the integral of equant, that's ln of the absolute value. Of 2 x plus tangent x. And we want to evaluate the result from 0 up to 5 divided by 6. So now let's go ahead and. Use the elements of integration and apply the evaluation theorem so we would have a of the absolute value of. Now equant can be expressed as 1 divided by cosine, so we would have l of the absolute value of 1 divided by cosine of pi divided by 6. Plus Tangents of pi divided by 6. And we're going to subtract ln of the absolute value of 1 divided by. Cosine of 0. Plus tangents of 0. What we want to do is simplify from here, so that'll be L and off. Now, cosine of pi divided by 6 is going to be the square root of 3 divided by 2, and if we take the reciprocal value of square root of 3. Divided by 2, that would be 2 divided by square root of 3. We're going to add tangent of 5 divided by 6, which is. 1 divided by a square root of 3. And we're subtracting eon of 1 divided by cosine of 0, that's 1 divided by 1, which is 1, plus tangent of 0, which is 0. Overall, that's eon of 1 which is 0, right? So we don't have that sum. Now let's go ahead and simplify, so we are going to get L and off. Notice that we can drop the absolute value because we have a sum of 2 positive numbers. And that would be Ellen of 3 divided by the square root of 3, which is square root of 3. And the square root of 3 can be written as 3 raises the power of 1/2 so that now we can apply the properties of logarithms and bring down the exponent that would be 1/2 ln of 3. Which corresponds to the answer of choice c. Thank you for watching.

In Exercises 139–142, find the length of each curve.141. y = ln(cos(x)) from x = 0 to x = π/4.

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In Exercises 139–142, find the length of each curve.
141. y = ln(cos(x)) from x = 0 to x = π/4.