Textbook Question
Solve the differential equation in Exercises 9–22.
10. (dy/dx) = x²√y, y > 0
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Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.6.96
Evaluate the integrals in Exercises 91–102.
96. ∫dy/((arcsin y)(1-y²))
Verified step by step guidance1
Recognize that the integral is of the form \(\int \frac{dy}{(\arcsin y)(1 - y^2)}\). Notice that the denominator contains \(1 - y^2\), which is related to the derivative of \(\arcsin y\).
Recall the derivative formula: \(\frac{d}{dy}(\arcsin y) = \frac{1}{\sqrt{1 - y^2}}\). This suggests a substitution involving \(\arcsin y\) might simplify the integral.
Let \(t = \arcsin y\). Then, differentiate both sides to find \(dy\) in terms of \(dt\): \(dy = \cos t \, dt\), since \(y = \sin t\) and \(dy/dt = \cos t\).
Rewrite the integral in terms of \(t\): substitute \(y = \sin t\) and \(dy = \cos t \, dt\). Also, express \(1 - y^2\) as \(1 - \sin^2 t = \cos^2 t\). The integral becomes \(\int \frac{\cos t \, dt}{t \cdot \cos^2 t} = \int \frac{dt}{t \cdot \cos t}\).
Now, the integral is \(\int \frac{dt}{t \cos t}\). From here, consider methods such as integration by parts or special functions to proceed, depending on the context or allowed techniques.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Integration Techniques Involving Inverse Trigonometric Functions
Integrals containing inverse trigonometric functions like arcsin y often require substitution or recognition of derivative forms. Understanding how to manipulate these functions and their derivatives is essential to simplify and evaluate the integral.
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Integrals Resulting in Inverse Trig Functions
Derivative of arcsin y
The derivative of arcsin y is 1/√(1 - y²). Recognizing this derivative helps in identifying substitution opportunities, especially when the integral contains expressions like 1/(1 - y²) or its powers.
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Substitution Method in Integration
Substitution involves changing variables to simplify an integral. For this problem, setting u = arcsin y transforms the integral into a more manageable form, allowing the integral to be expressed in terms of u and evaluated more easily.
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Related Practice
Textbook Question
Indeterminate Powers and Products
Find the limits in Exercises 53–68.
60. lim (x → 0) (e^x + x)^(1/x)
Textbook Question
Evaluate the integrals in Exercises 53–76.
75. ∫y dy/√(1-y^4)
Textbook Question
22. The function ln x grows slower than any polynomial Show that ln(x) grows slower as x→∞ than any nonconstant polynomial.
Textbook Question
Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.
f(x) = x⁵
Textbook Question
In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.
119. y = (sin x)ˣ