7–84. Evaluate the following integrals.
30. ∫ from 5/2 to 5√3/2 [1 / (v² √(25 - v²))] dv

Verified step by step guidance

1. Step 1: Recognize the integral's structure. The integrand resembles the form of a trigonometric substitution problem, specifically involving the square root of a difference of squares, √(a² - v²). This suggests using a substitution based on the Pythagorean identity.
2. Step 2: Set up the substitution. Let v = 5 sin(θ), where 5 is the square root of the constant term 25. Then, dv = 5 cos(θ) dθ, and √(25 - v²) becomes √(25 - 25 sin²(θ)) = √(25(1 - sin²(θ))) = 5 cos(θ).
3. Step 3: Change the limits of integration. When v = 5/2, solve for θ using v = 5 sin(θ): sin(θ) = (5/2)/5 = 1/2, so θ = π/6. When v = 5√3/2, solve for θ: sin(θ) = (5√3/2)/5 = √3/2, so θ = π/3.
4. Step 4: Rewrite the integral in terms of θ. Substitute v = 5 sin(θ), dv = 5 cos(θ) dθ, and √(25 - v²) = 5 cos(θ) into the integral. The integrand simplifies to ∫ from π/6 to π/3 [1 / (25 sin²(θ) * 5 cos(θ))] * 5 cos(θ) dθ, which further simplifies to ∫ from π/6 to π/3 [1 / 25 sin²(θ)] dθ.
5. Step 5: Simplify and evaluate the integral. Factor out constants and use trigonometric identities or integration techniques to solve ∫ from π/6 to π/3 [1 / sin²(θ)] dθ. Recall that 1 / sin²(θ) can be expressed as csc²(θ), whose integral is -cot(θ). Apply the limits of integration to find the result.

Key Concepts

Definite Integrals

A definite integral calculates the accumulation of a quantity, represented as the area under a curve, between two specified limits. In this case, the integral is evaluated from 5/2 to 5√3/2, which means we are interested in the net area between the curve of the function and the x-axis over this interval.

Integration Techniques

To evaluate integrals like the one presented, various techniques may be employed, such as substitution or trigonometric identities. The integrand, 1 / (v² √(25 - v²)), suggests that a trigonometric substitution might simplify the expression, particularly since it involves a square root of a difference.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, stating that if a function is continuous on an interval, the integral of its derivative over that interval equals the difference in the values of the function at the endpoints. This theorem allows us to evaluate definite integrals by finding an antiderivative of the integrand and applying the limits of integration.