Chapter 7: Problem 85
Evaluate the integrals in Exercises \(81-90\) $$ \int \frac{d y}{\left(\tan ^{-1} y\right)\left(1+y^{2}\right)} $$
Short Answer
Expert verified
The integral evaluates to \( \ln |\tan^{-1} y| + C \).
Step by step solution
01
Identify the Integral
The given integral is \( \int \frac{dy}{(\tan^{-1} y)(1+y^2)} \). Our task is to find a suitable substitution or method to simplify and evaluate it.
02
Substitution Choice
We will use the substitution \( u = \tan^{-1} y \). This implies that \( y = \tan(u) \) and \( dy = \sec^2(u) \, du \). The identity \( 1 + y^2 = \sec^2(u) \) becomes useful for simplifying.
03
Substitute Into the Integral
Replacing \( y \) and \( dy \) in the integral, we have:\[\int \frac{dy}{(\tan^{-1} y)(1+y^2)} = \int \frac{\sec^2(u) \, du}{u \cdot \sec^2(u)}.\]Simplifying, this becomes \( \int \frac{du}{u} \).
04
Evaluate the Simplified Integral
The integral \( \int \frac{du}{u} \) is a standard integral, which evaluates to \( \ln |u| + C \), where \( C \) is the integration constant.
05
Back-substitute the Original Variable
Since we used \( u = \tan^{-1} y \) for substitution, substitute back to get the solution in terms of \( y \):\[\ln |\tan^{-1} y| + C.\]
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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 used to calculate the area under a curve from a specific point to another. They are different from indefinite integrals, which do not have specified limits and result in a family of functions. In the context of definite integrals, the integral limits are noted, usually written as \[ \int_{a}^{b} f(x) \, dx \].The process involves finding an antiderivative of the function and then applying the fundamental theorem of calculus:
- First, find the antiderivative, also known as the indefinite integral.
- Next, evaluate the antiderivative at both endpoints of the interval.
- Finally, subtract the value at the lower bound from the value at the upper bound.
Trigonometric Substitution
Trigonometric substitution is a technique to simplify integration, specifically useful for dealing with integrals involving square roots. This strategy replaces variables with trigonometric functions. It can often reduce a complex integral into something easier to manage.When you consider substitutions like
- \( y = \tan(u) \)
- \( dy = \sec^2(u) \, du \)
Antiderivatives
In calculus, an antiderivative, often called an indefinite integral, is a function whose derivative is the original function. When you're faced with an integral like \[ \int \frac{du}{u} \],the goal is to find a function \( F(u) \) such that \[ F'(u) = \frac{1}{u} \].The integral of \( \frac{1}{u} \) is a classic example, resulting in the natural logarithm: \[ F(u) = \ln|u| + C \],where \( C \) represents the constant of integration.Antiderivatives are critical for solving differential equations and evaluating definite integrals. They allow us to work backward from a rate of change to determine the original function, like reversing the process of taking a derivative.For instance, in the given exercise, after a substitution strategy simplified the integral, finding the antiderivative led to the solution. This showcases how antiderivatives provide the means to solve complex integration problems efficiently.