Evaluate the integrals in Exercises 53–76.
59. ∫(from 0 to 1)4ds/√(4-s²)

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Recognize that the integral is of the form \(\int \frac{4}{\sqrt{4 - s^2}} \, ds\) from \(0\) to \(1\). This resembles the integral of a function involving \(\sqrt{a^2 - s^2}\), where \(a=2\) because \(4 = 2^2\).

Recall the standard integral formula: \(\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C\). Since the numerator here is \(4\), factor it out to write the integral as \(4 \int \frac{1}{\sqrt{4 - s^2}} \, ds\).

Apply the formula to the integral inside: \(\int \frac{1}{\sqrt{4 - s^2}} \, ds = \arcsin\left(\frac{s}{2}\right) + C\).

Evaluate the definite integral by substituting the limits \(s=0\) and \(s=1\) into \(4 \arcsin\left(\frac{s}{2}\right)\), i.e., compute \(4 \left[ \arcsin\left(\frac{1}{2}\right) - \arcsin(0) \right]\).

Use known values of the arcsine function to express the result in terms of radians (e.g., \(\arcsin(0) = 0\) and \(\arcsin(\frac{1}{2}) = \frac{\pi}{6}\)), but do not calculate the final numeric value as per instructions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integral

A definite integral calculates the net area under a curve between two specified limits. It is represented as ∫ from a to b of f(x) dx, where a and b are the lower and upper bounds. Evaluating a definite integral involves finding the antiderivative and then applying the Fundamental Theorem of Calculus.

Integration of Functions Involving Square Roots

Integrals containing expressions like √(a² - x²) often require trigonometric substitution or recognition of standard integral forms. These integrals relate to inverse trigonometric functions, such as arcsin or arccos, which simplify the evaluation process.

Fundamental Theorem of Calculus

This theorem connects differentiation and integration, stating that if F is an antiderivative of f on [a, b], then ∫ from a to b of f(x) dx = F(b) - F(a). It allows evaluation of definite integrals by computing the difference of antiderivative values at the bounds.

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