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Chapter 8: Problem 12

Evaluate the integrals. $$\int \frac{\sqrt{y^{2}-25}}{y^{3}} d y, \quad y>5$$

Short Answer

Expert verified
Use trigonometric substitution and back-substitute to solve.

Step by step solution

01

Identify the Integral Type

The given integral is \( \int \frac{\sqrt{y^{2}-25}}{y^{3}} \, dy \). It involves a square root in the numerator and a polynomial in the denominator, indicating that a trigonometric substitution might simplify the expression.
02

Apply Trigonometric Substitution

For integrals involving \( \sqrt{y^2 - a^2} \), use the substitution \( y = a \sec \theta \), where \( a = 5 \). So, set \( y = 5 \sec \theta \), hence \( dy = 5 \sec \theta \tan \theta \, d\theta \). Substitute these into the integral.
03

Perform the Substitution

The square root \( \sqrt{y^2 - 25} = \sqrt{(5 \sec \theta)^2 - 25} = 5 \tan \theta \). Substitute \( y = 5 \sec \theta \) and \( dy = 5 \sec \theta \tan \theta \, d\theta \) into the integral: \[ \int \frac{5 \tan \theta}{(5 \sec \theta)^3} (5 \sec \theta \tan \theta) \, d\theta \].This simplifies to \[ \int \frac{25 \tan^2 \theta}{125 \sec^3 \theta} \, d\theta = \int \frac{\tan^2 \theta}{5 \sec^3 \theta} \, d\theta \].
04

Simplify Using Trigonometric Identities

Recall that \( \sec^2 \theta = 1 + \tan^2 \theta \), so \( \tan^2 \theta = \sec^2 \theta - 1 \). Thus, the integral becomes \[ \int \frac{\sec^2 \theta - 1}{5 \sec^3 \theta} \, d\theta = \int \frac{1}{5 \sec \theta} \, d\theta - \int \frac{1}{5 \sec^3 \theta} \, d\theta \].
05

Integrate Simplified Expression

The first integral, \( \int \frac{1}{5 \sec \theta} \, d\theta = \int \frac{\cos \theta}{5} \, d\theta = \frac{1}{5} \int \cos \theta \, d\theta \), integrates to \( \frac{1}{5} \sin \theta + C_1 \).The second integral, \( \int \frac{1}{5 \sec^3 \theta} \, d\theta = \int \frac{\cos^3 \theta}{5} \, d\theta \), is more complex. Use integration by parts or another trigonometric identity to integrate it if necessary. For now, focus on easily integrable parts.
06

Back-Substitute to Original Variable

Since \( y = 5 \sec \theta \), \( \sec \theta = \frac{y}{5} \) and \( \cos \theta = \frac{5}{y} \). Knowing that \( \tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\frac{y^2}{25} - 1} = \frac{\sqrt{y^2 - 25}}{5} \), substitute back:\[ \frac{1}{5} \sin \theta = \frac{1}{5} \frac{\sqrt{y^2 - 25}}{y} \].Evaluate the leftover integral as needed or apply other methods as required to simplify.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integrals
Definite integrals are a fundamental concept in calculus. They allow us to calculate the total accumulation of quantities, like area under curves. A definite integral has limits, indicating where the accumulation starts and ends. We denote these with numbers at the top and bottom of the integral sign.
In contrast, an indefinite integral calculates an antiderivative without any limits.
  • For example, to find the area between the curve and the x-axis from 1 to 3, you'd compute \(\int_1^3 f(x) \, dx\).
  • This gives a precise "sum" of the area.
In our exercise, though, the focus is on finding the antiderivative, without numerical limits. After simplifying our expression using a trigonometric substitution, it would be possible to evaluate the integral from specific bounds if needed.
Trigonometric Identities
Trigonometric identities are equations true for all angles. They relate the trigonometric functions like sine, cosine, and tangent to one another. These identities are powerful tools for simplifying expressions.
  • In our problem, by recognizing \( y^2 - 25 \) paired with \( y = 5 \sec \theta \), we use the identity \( \sec^2 \theta = 1 + \tan^2 \theta \).
  • This helps us express the complicated square root in a simpler trigonometric form.
This substitution turns our integral into expressions that are easier to integrate.
Knowing these identities allows us to reframe complex expressions into more manageable forms, helping at every step of the integration process.
Square Root Functions
Square root functions often appear in calculus, especially within integrals involving differences like \( y^2 - 25 \). These functions can be tricky during integration.
To tackle these effectively, we use specific substitutions to simplify the square root part.
  • For example, by substituting \( y = 5 \sec \theta \), we express the square root \( \sqrt{y^2 - 25} \) in terms of \( \tan \theta \).
  • This "transforms" the function and eliminates the need to deal with complicated algebraic expressions under the root directly.
Understanding how and why we make these substitutions allows us to tackle a broader range of problems involving square roots in integrals.
They convert complex integration problems into solvable trigonometric integrals.

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Most popular questions from this chapter

Manufacturing time The assembly time in minutes for a component at an electronic manufacturing plant is normally distributed with a mean of \(\mu=55\) and standard deviation \(\sigma=4 .\) What is the probability that a component will be made in less than one hour?

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Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{\ln 2} x^{-2} e^{-1 / x} d x$$

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