Problem 15
Question
Use the table of integrals at the back of the book to evaluate the integrals. \(\int \sqrt{25-p^{2}} d p\)
Step-by-Step Solution
Verified Answer
\( \frac{1}{2} (p \sqrt{25 - p^2} + 25 \sin^{-1}(\frac{p}{5})) + C \)
1Step 1: Recognize the Integral Form
The integral \( \int \sqrt{25 - p^2} \, dp \) resembles the form of a standard trigonometric integral \( \int \sqrt{a^2 - x^2} \, dx \). Here, \( a = 5 \) and \( x = p \).
2Step 2: Use the Table of Integrals
Refer to the integral table, where it lists the standard form \( \int \sqrt{a^2 - x^2} \, dx = \frac{1}{2}(x\sqrt{a^2-x^2} + a^2 \sin^{-1}(\frac{x}{a})) + C \).
3Step 3: Substitute and Integrate
Substitute \( a = 5 \) and \( x = p \) into the formula. Thus, the integral becomes: \[ \frac{1}{2} (p \sqrt{25 - p^2} + 25 \sin^{-1}(\frac{p}{5})) + C \] where \( C \) is the constant of integration.
Key Concepts
Understanding Trigonometric IntegralsUtilizing the Table of IntegralsThe Role of the Constant of Integration
Understanding Trigonometric Integrals
Trigonometric integrals are a type of integral that involves trigonometric functions such as sine, cosine, and tangent. They often appear in mathematical problems when we deal with functions involving angles or periodic phenomena.
In the case of the integral \( \int \sqrt{25-p^{2}} \, dp \), although it doesn't explicitly involve trigonometric functions from the start, it resembles forms that can be linked to trigonometric identities. This connection arises because expressions like \( \sqrt{a^2 - x^2} \) can be related to the Pythagorean identity \( \sin^2{\theta} + \cos^2{\theta} = 1 \). By recognizing such patterns, integration becomes more manageable.
Using trigonometric substitutions is a powerful tool that allows the transformation of complex integrals into simpler forms by substituting variables to match standard trigonometric identities or derivatives.
In the case of the integral \( \int \sqrt{25-p^{2}} \, dp \), although it doesn't explicitly involve trigonometric functions from the start, it resembles forms that can be linked to trigonometric identities. This connection arises because expressions like \( \sqrt{a^2 - x^2} \) can be related to the Pythagorean identity \( \sin^2{\theta} + \cos^2{\theta} = 1 \). By recognizing such patterns, integration becomes more manageable.
Using trigonometric substitutions is a powerful tool that allows the transformation of complex integrals into simpler forms by substituting variables to match standard trigonometric identities or derivatives.
Utilizing the Table of Integrals
The table of integrals is an invaluable resource for students and professionals alike, as it provides a handy collection of integral forms and solutions. It's especially beneficial when dealing with integrals that do not lend themselves to easy analytical solutions.
For the integral \( \int \sqrt{25 - p^2} \, dp \), we utilized a listed form \( \int \sqrt{a^2 - x^2} \ dx = \frac{1}{2}(x\sqrt{a^2-x^2} + a^2 \sin^{-1}(\frac{x}{a})) + C \), which simplifies the integration process by providing a direct formula once the parameters are identified. This saves time and effort by reducing complex algebra and calculus steps to straightforward substitutions.
For the integral \( \int \sqrt{25 - p^2} \, dp \), we utilized a listed form \( \int \sqrt{a^2 - x^2} \ dx = \frac{1}{2}(x\sqrt{a^2-x^2} + a^2 \sin^{-1}(\frac{x}{a})) + C \), which simplifies the integration process by providing a direct formula once the parameters are identified. This saves time and effort by reducing complex algebra and calculus steps to straightforward substitutions.
- This approach is especially useful in educational settings, where understanding integral forms through repetition enhances both comprehension and efficiency.
The Role of the Constant of Integration
In integral calculus, when finding antiderivatives, the constant of integration \( C \) plays a crucial role. It acknowledges the fact that indefinite integrals represent a family of functions rather than a single solution.
When an integral is evaluated, such as \( \int \sqrt{25 - p^2} \, dp \), adding \( C \) at the end of the solution signifies that other functions exist that differ by just this constant. These different functions differ by a vertical shift on a graph. Whether the context is purely academic or applied, never neglect the constant of integration.
When an integral is evaluated, such as \( \int \sqrt{25 - p^2} \, dp \), adding \( C \) at the end of the solution signifies that other functions exist that differ by just this constant. These different functions differ by a vertical shift on a graph. Whether the context is purely academic or applied, never neglect the constant of integration.
- This constant becomes essential when initial conditions or specific points are given to find the unique solution among the family of functions.
Other exercises in this chapter
Problem 15
Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{1} \frac{\theta+1}{\sqrt{\theta^{2}+2 \theta}} d \theta $$
View solution Problem 15
Evaluate the integrals in Exercises \(15-22\). $$ \int_{0}^{2 \pi} \sqrt{\frac{1-\cos x}{2}} d x $$
View solution Problem 15
Evaluate the integrals. \(\int x^{5} e^{x} d x\)
View solution Problem 15
In Exercises \(9-16,\) express the integrands as a sum of partial fractions and evaluate the integrals. $$ \int \frac{d t}{t^{3}+t^{2}-2 t} $$
View solution