91Ó°ÊÓ

When dealing with functions like \( f(x) = \ln(\sin x) \), derivatives are crucial, especially for computing arc lengths. The derivative of a function gives us a way to understand how the function changes at any given point. For \( f(x) = \ln(\sin x) \), finding the derivative requires the use of the chain rule.

• The chain rule helps differentiate composed functions, like \( \ln(\sin x) \).
• To apply it: differentiate the outer function \( \ln u \) to get \( \frac{1}{u} \), then multiply by the derivative of the inner function \( \sin x \), which is \( \cos x \).

This process gives us \( f'(x) = \frac{1}{\sin x} \cdot \cos x = \cot x \), indicating that the rate of change of the logarithm of sine is essentially the cotangent of \( x \). Understanding this derivative is key when integrating to find the arc length.

Trigonometric identities are vital tools in simplifying and solving problems involving trigonometric functions. In the context of finding arc lengths, they help simplify expressions before integration. A significant identity used here is:

• \( 1 + \cot^2(x) = \csc^2(x) \)

This identity transforms the expression under the square root of the arc length formula. The original form \( \sqrt{1+\cot^2(x)} \) becomes \( \csc(x) \). This simplification streamlines integration. It's crucial because it changes a potentially complex integral into a form where the integral of \( \csc(x) \) has a known antiderivative, making the whole process more manageable. Utilizing such identities turns challenging problems into ones that are easier to handle, providing a clearer path to the solution.

The definite integral is a concept used to find the total accumulation of a quantity, such as the arc length over a specific interval. In this problem, the arc length from \( \pi/6 \) to \( \pi/2 \) is calculated using a definite integral.

• We start with the integral \( \int_{\pi/6}^{\pi/2} \csc(x) \, dx \), simplified using trigonometric identities.
• The antiderivative of \( \csc(x) \) is \( -\ln|\csc(x) + \cot(x)| \).

Evaluating this from \( \pi/6 \) to \( \pi/2 \) involves substituting these bounds into the antiderivative.The process includes calculating the value at each limit and subtracting:

• At \( x = \pi/2 \), the function simplifies to 0, contributing no arc length.
• At \( x = \pi/6 \), it calculates \( -\ln|2 + \sqrt{3}| \), a natural logarithm involving the cosecant and cotangent values at \( \pi/6 \).

This final computation gives the arc length as \( \ln(2 + \sqrt{3}) \), illustrating how the definite integral encapsulates the entire arc of the function over the given interval.